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Converting a string into numbers


MIPS: determine if a list of test scores are pass/failMIPS: Check how many alphabetic characters are in a stringHow do I find the mode digit with the least number of instruction counts?MIPS assembly language printing out each int of array line by lineMultiplying and Combining MatricesPrinting 10 characters from a String at a time in MIPSHow to remove a comma from string inputMIPS code doesn't print strings loaded from memoryMIPS reverse order of Capital Letters in each word in stringMIPS : parsing and modifying a string






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








0















# replacing all digits in a string with their complement in 9.
.data
string: .asciiz "123471863"

.text
main:
# load string's 1st address into the memory
la $a0, string

# initialize the loop-counter
li $t0, 0
li $t1, 9 # complement envelope for later use

# start the loop
start_loop:
lb $t2, ($a0) # Take one character from string

# loop termination condition
beq $t2, $zero, end_loop # terminate if null-value found

subi $t2, $t2, 48 # convert it to a digit
sub $t2, $t1, $t2 # apply complement to $t2
sw $t2,($a0) # restore the string-byte content

addi $a0, $a0, 1 # go to next string-byte
addi $t0, $t0, 1 # increment loop-counter

j start_loop
end_loop:

# print string
la $a0, string # load 1st address of the string
li $v0, 4 # syscall for string print
syscall

move $a0, $t0 # load 1st address of the string
li $v0, 1 # syscall for string print
syscall

# exit program
li $v0, 10
syscall


The program isn't working as expected. After 1st iteration, the $a0 register isn't giving the correct value. Apparently, sw $t2,($a0) is destroying the original address.



How can I get over this issue?










share|improve this question
























  • "null", in the sense of the comment, is a zero value. For handling strings with embedded zero bytes (i.e. strings where the characters can be any byte value) you need to use a different way of encoding strings, the most common being to prefix the string with a length value.

    – Josh Greifer
    Mar 24 at 16:28












  • @JoshGreifer, how to do that?

    – user366312
    Mar 24 at 16:32











  • Well, that's another question which you can post here - but I suggest you first look for questions here on "pascal strings"

    – Josh Greifer
    Mar 24 at 16:34












  • @JoshGreifer, "that's another question which you can post here" --- this is not another question. On the basis of your comment, this is already part of this question.

    – user366312
    Mar 24 at 16:36


















0















# replacing all digits in a string with their complement in 9.
.data
string: .asciiz "123471863"

.text
main:
# load string's 1st address into the memory
la $a0, string

# initialize the loop-counter
li $t0, 0
li $t1, 9 # complement envelope for later use

# start the loop
start_loop:
lb $t2, ($a0) # Take one character from string

# loop termination condition
beq $t2, $zero, end_loop # terminate if null-value found

subi $t2, $t2, 48 # convert it to a digit
sub $t2, $t1, $t2 # apply complement to $t2
sw $t2,($a0) # restore the string-byte content

addi $a0, $a0, 1 # go to next string-byte
addi $t0, $t0, 1 # increment loop-counter

j start_loop
end_loop:

# print string
la $a0, string # load 1st address of the string
li $v0, 4 # syscall for string print
syscall

move $a0, $t0 # load 1st address of the string
li $v0, 1 # syscall for string print
syscall

# exit program
li $v0, 10
syscall


The program isn't working as expected. After 1st iteration, the $a0 register isn't giving the correct value. Apparently, sw $t2,($a0) is destroying the original address.



How can I get over this issue?










share|improve this question
























  • "null", in the sense of the comment, is a zero value. For handling strings with embedded zero bytes (i.e. strings where the characters can be any byte value) you need to use a different way of encoding strings, the most common being to prefix the string with a length value.

    – Josh Greifer
    Mar 24 at 16:28












  • @JoshGreifer, how to do that?

    – user366312
    Mar 24 at 16:32











  • Well, that's another question which you can post here - but I suggest you first look for questions here on "pascal strings"

    – Josh Greifer
    Mar 24 at 16:34












  • @JoshGreifer, "that's another question which you can post here" --- this is not another question. On the basis of your comment, this is already part of this question.

    – user366312
    Mar 24 at 16:36














0












0








0








# replacing all digits in a string with their complement in 9.
.data
string: .asciiz "123471863"

.text
main:
# load string's 1st address into the memory
la $a0, string

# initialize the loop-counter
li $t0, 0
li $t1, 9 # complement envelope for later use

# start the loop
start_loop:
lb $t2, ($a0) # Take one character from string

# loop termination condition
beq $t2, $zero, end_loop # terminate if null-value found

subi $t2, $t2, 48 # convert it to a digit
sub $t2, $t1, $t2 # apply complement to $t2
sw $t2,($a0) # restore the string-byte content

addi $a0, $a0, 1 # go to next string-byte
addi $t0, $t0, 1 # increment loop-counter

j start_loop
end_loop:

# print string
la $a0, string # load 1st address of the string
li $v0, 4 # syscall for string print
syscall

move $a0, $t0 # load 1st address of the string
li $v0, 1 # syscall for string print
syscall

# exit program
li $v0, 10
syscall


The program isn't working as expected. After 1st iteration, the $a0 register isn't giving the correct value. Apparently, sw $t2,($a0) is destroying the original address.



How can I get over this issue?










share|improve this question
















# replacing all digits in a string with their complement in 9.
.data
string: .asciiz "123471863"

.text
main:
# load string's 1st address into the memory
la $a0, string

# initialize the loop-counter
li $t0, 0
li $t1, 9 # complement envelope for later use

# start the loop
start_loop:
lb $t2, ($a0) # Take one character from string

# loop termination condition
beq $t2, $zero, end_loop # terminate if null-value found

subi $t2, $t2, 48 # convert it to a digit
sub $t2, $t1, $t2 # apply complement to $t2
sw $t2,($a0) # restore the string-byte content

addi $a0, $a0, 1 # go to next string-byte
addi $t0, $t0, 1 # increment loop-counter

j start_loop
end_loop:

# print string
la $a0, string # load 1st address of the string
li $v0, 4 # syscall for string print
syscall

move $a0, $t0 # load 1st address of the string
li $v0, 1 # syscall for string print
syscall

# exit program
li $v0, 10
syscall


The program isn't working as expected. After 1st iteration, the $a0 register isn't giving the correct value. Apparently, sw $t2,($a0) is destroying the original address.



How can I get over this issue?







assembly mips






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 24 at 16:53







user366312

















asked Mar 24 at 16:18









user366312user366312

4,03947162321




4,03947162321












  • "null", in the sense of the comment, is a zero value. For handling strings with embedded zero bytes (i.e. strings where the characters can be any byte value) you need to use a different way of encoding strings, the most common being to prefix the string with a length value.

    – Josh Greifer
    Mar 24 at 16:28












  • @JoshGreifer, how to do that?

    – user366312
    Mar 24 at 16:32











  • Well, that's another question which you can post here - but I suggest you first look for questions here on "pascal strings"

    – Josh Greifer
    Mar 24 at 16:34












  • @JoshGreifer, "that's another question which you can post here" --- this is not another question. On the basis of your comment, this is already part of this question.

    – user366312
    Mar 24 at 16:36


















  • "null", in the sense of the comment, is a zero value. For handling strings with embedded zero bytes (i.e. strings where the characters can be any byte value) you need to use a different way of encoding strings, the most common being to prefix the string with a length value.

    – Josh Greifer
    Mar 24 at 16:28












  • @JoshGreifer, how to do that?

    – user366312
    Mar 24 at 16:32











  • Well, that's another question which you can post here - but I suggest you first look for questions here on "pascal strings"

    – Josh Greifer
    Mar 24 at 16:34












  • @JoshGreifer, "that's another question which you can post here" --- this is not another question. On the basis of your comment, this is already part of this question.

    – user366312
    Mar 24 at 16:36

















"null", in the sense of the comment, is a zero value. For handling strings with embedded zero bytes (i.e. strings where the characters can be any byte value) you need to use a different way of encoding strings, the most common being to prefix the string with a length value.

– Josh Greifer
Mar 24 at 16:28






"null", in the sense of the comment, is a zero value. For handling strings with embedded zero bytes (i.e. strings where the characters can be any byte value) you need to use a different way of encoding strings, the most common being to prefix the string with a length value.

– Josh Greifer
Mar 24 at 16:28














@JoshGreifer, how to do that?

– user366312
Mar 24 at 16:32





@JoshGreifer, how to do that?

– user366312
Mar 24 at 16:32













Well, that's another question which you can post here - but I suggest you first look for questions here on "pascal strings"

– Josh Greifer
Mar 24 at 16:34






Well, that's another question which you can post here - but I suggest you first look for questions here on "pascal strings"

– Josh Greifer
Mar 24 at 16:34














@JoshGreifer, "that's another question which you can post here" --- this is not another question. On the basis of your comment, this is already part of this question.

– user366312
Mar 24 at 16:36






@JoshGreifer, "that's another question which you can post here" --- this is not another question. On the basis of your comment, this is already part of this question.

– user366312
Mar 24 at 16:36













1 Answer
1






active

oldest

votes


















1














There is no problem to differentiate a null and a '0'. null is 0, while '' is 48.



Your test



 beq $t2, $zero, end_loop # terminate if null-value found


is perfectly correct and will detect the end of string.



What is incorrect is you algorithm.



A way to complement a number in C, would be :



while(c=*str)
c=c-'0' ; // transform the number to integer
c=9-c; // complement it
c += '0'; // add 48 to turn it back to a character
str++;



You are missing the last conversion to character.



If you change



 sub $t2, $t1, $t2 # apply complement to $t2


to



 sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


all should work.



Alternatively, ou can simplify your algorithm and remark that the computation c=9-(c-48)+48 is equivalent to c=105-c. Add before start_loop



 li $t4 105 ## 


and replace the three lines



 subi $t2, $t2, 48 # convert it to a digit
sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


by



 sub $t2,$t4,$t2 # complement to 9 directly on char representing the digit





share|improve this answer























  • li $t4 105 ## ?

    – user366312
    Mar 24 at 17:17











  • Ooops, a comma is missing. li $t4,105 The problem is that you can subtract an immediate from a register, but not subtract a register from an immediate in an instruction. So you put the immediate 105 in a register before the loop to perform the subtraction in an instruction.

    – Alain Merigot
    Mar 24 at 17:25











  • If you change sub $t2, $t1, $t2 # apply complement to $t2 to sub $t2, $t1, $t2 # apply complement to $t2 addi $t2, $t2, 48 all should work. --- nope. doesn't work.

    – user366312
    Mar 24 at 17:30











  • Onec corrected, your algorithm really seems correct, but I do not have time to test it. Maybe the problem comes from branches. Are your branches delayed ?

    – Alain Merigot
    Mar 24 at 17:37











  • Are your branches delayed? --- nope.

    – user366312
    Mar 24 at 17:47












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














There is no problem to differentiate a null and a '0'. null is 0, while '' is 48.



Your test



 beq $t2, $zero, end_loop # terminate if null-value found


is perfectly correct and will detect the end of string.



What is incorrect is you algorithm.



A way to complement a number in C, would be :



while(c=*str)
c=c-'0' ; // transform the number to integer
c=9-c; // complement it
c += '0'; // add 48 to turn it back to a character
str++;



You are missing the last conversion to character.



If you change



 sub $t2, $t1, $t2 # apply complement to $t2


to



 sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


all should work.



Alternatively, ou can simplify your algorithm and remark that the computation c=9-(c-48)+48 is equivalent to c=105-c. Add before start_loop



 li $t4 105 ## 


and replace the three lines



 subi $t2, $t2, 48 # convert it to a digit
sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


by



 sub $t2,$t4,$t2 # complement to 9 directly on char representing the digit





share|improve this answer























  • li $t4 105 ## ?

    – user366312
    Mar 24 at 17:17











  • Ooops, a comma is missing. li $t4,105 The problem is that you can subtract an immediate from a register, but not subtract a register from an immediate in an instruction. So you put the immediate 105 in a register before the loop to perform the subtraction in an instruction.

    – Alain Merigot
    Mar 24 at 17:25











  • If you change sub $t2, $t1, $t2 # apply complement to $t2 to sub $t2, $t1, $t2 # apply complement to $t2 addi $t2, $t2, 48 all should work. --- nope. doesn't work.

    – user366312
    Mar 24 at 17:30











  • Onec corrected, your algorithm really seems correct, but I do not have time to test it. Maybe the problem comes from branches. Are your branches delayed ?

    – Alain Merigot
    Mar 24 at 17:37











  • Are your branches delayed? --- nope.

    – user366312
    Mar 24 at 17:47
















1














There is no problem to differentiate a null and a '0'. null is 0, while '' is 48.



Your test



 beq $t2, $zero, end_loop # terminate if null-value found


is perfectly correct and will detect the end of string.



What is incorrect is you algorithm.



A way to complement a number in C, would be :



while(c=*str)
c=c-'0' ; // transform the number to integer
c=9-c; // complement it
c += '0'; // add 48 to turn it back to a character
str++;



You are missing the last conversion to character.



If you change



 sub $t2, $t1, $t2 # apply complement to $t2


to



 sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


all should work.



Alternatively, ou can simplify your algorithm and remark that the computation c=9-(c-48)+48 is equivalent to c=105-c. Add before start_loop



 li $t4 105 ## 


and replace the three lines



 subi $t2, $t2, 48 # convert it to a digit
sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


by



 sub $t2,$t4,$t2 # complement to 9 directly on char representing the digit





share|improve this answer























  • li $t4 105 ## ?

    – user366312
    Mar 24 at 17:17











  • Ooops, a comma is missing. li $t4,105 The problem is that you can subtract an immediate from a register, but not subtract a register from an immediate in an instruction. So you put the immediate 105 in a register before the loop to perform the subtraction in an instruction.

    – Alain Merigot
    Mar 24 at 17:25











  • If you change sub $t2, $t1, $t2 # apply complement to $t2 to sub $t2, $t1, $t2 # apply complement to $t2 addi $t2, $t2, 48 all should work. --- nope. doesn't work.

    – user366312
    Mar 24 at 17:30











  • Onec corrected, your algorithm really seems correct, but I do not have time to test it. Maybe the problem comes from branches. Are your branches delayed ?

    – Alain Merigot
    Mar 24 at 17:37











  • Are your branches delayed? --- nope.

    – user366312
    Mar 24 at 17:47














1












1








1







There is no problem to differentiate a null and a '0'. null is 0, while '' is 48.



Your test



 beq $t2, $zero, end_loop # terminate if null-value found


is perfectly correct and will detect the end of string.



What is incorrect is you algorithm.



A way to complement a number in C, would be :



while(c=*str)
c=c-'0' ; // transform the number to integer
c=9-c; // complement it
c += '0'; // add 48 to turn it back to a character
str++;



You are missing the last conversion to character.



If you change



 sub $t2, $t1, $t2 # apply complement to $t2


to



 sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


all should work.



Alternatively, ou can simplify your algorithm and remark that the computation c=9-(c-48)+48 is equivalent to c=105-c. Add before start_loop



 li $t4 105 ## 


and replace the three lines



 subi $t2, $t2, 48 # convert it to a digit
sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


by



 sub $t2,$t4,$t2 # complement to 9 directly on char representing the digit





share|improve this answer













There is no problem to differentiate a null and a '0'. null is 0, while '' is 48.



Your test



 beq $t2, $zero, end_loop # terminate if null-value found


is perfectly correct and will detect the end of string.



What is incorrect is you algorithm.



A way to complement a number in C, would be :



while(c=*str)
c=c-'0' ; // transform the number to integer
c=9-c; // complement it
c += '0'; // add 48 to turn it back to a character
str++;



You are missing the last conversion to character.



If you change



 sub $t2, $t1, $t2 # apply complement to $t2


to



 sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


all should work.



Alternatively, ou can simplify your algorithm and remark that the computation c=9-(c-48)+48 is equivalent to c=105-c. Add before start_loop



 li $t4 105 ## 


and replace the three lines



 subi $t2, $t2, 48 # convert it to a digit
sub $t2, $t1, $t2 # apply complement to $t2
addi $t2, $t2, 48


by



 sub $t2,$t4,$t2 # complement to 9 directly on char representing the digit






share|improve this answer












share|improve this answer



share|improve this answer










answered Mar 24 at 17:11









Alain MerigotAlain Merigot

5,1992922




5,1992922












  • li $t4 105 ## ?

    – user366312
    Mar 24 at 17:17











  • Ooops, a comma is missing. li $t4,105 The problem is that you can subtract an immediate from a register, but not subtract a register from an immediate in an instruction. So you put the immediate 105 in a register before the loop to perform the subtraction in an instruction.

    – Alain Merigot
    Mar 24 at 17:25











  • If you change sub $t2, $t1, $t2 # apply complement to $t2 to sub $t2, $t1, $t2 # apply complement to $t2 addi $t2, $t2, 48 all should work. --- nope. doesn't work.

    – user366312
    Mar 24 at 17:30











  • Onec corrected, your algorithm really seems correct, but I do not have time to test it. Maybe the problem comes from branches. Are your branches delayed ?

    – Alain Merigot
    Mar 24 at 17:37











  • Are your branches delayed? --- nope.

    – user366312
    Mar 24 at 17:47


















  • li $t4 105 ## ?

    – user366312
    Mar 24 at 17:17











  • Ooops, a comma is missing. li $t4,105 The problem is that you can subtract an immediate from a register, but not subtract a register from an immediate in an instruction. So you put the immediate 105 in a register before the loop to perform the subtraction in an instruction.

    – Alain Merigot
    Mar 24 at 17:25











  • If you change sub $t2, $t1, $t2 # apply complement to $t2 to sub $t2, $t1, $t2 # apply complement to $t2 addi $t2, $t2, 48 all should work. --- nope. doesn't work.

    – user366312
    Mar 24 at 17:30











  • Onec corrected, your algorithm really seems correct, but I do not have time to test it. Maybe the problem comes from branches. Are your branches delayed ?

    – Alain Merigot
    Mar 24 at 17:37











  • Are your branches delayed? --- nope.

    – user366312
    Mar 24 at 17:47

















li $t4 105 ## ?

– user366312
Mar 24 at 17:17





li $t4 105 ## ?

– user366312
Mar 24 at 17:17













Ooops, a comma is missing. li $t4,105 The problem is that you can subtract an immediate from a register, but not subtract a register from an immediate in an instruction. So you put the immediate 105 in a register before the loop to perform the subtraction in an instruction.

– Alain Merigot
Mar 24 at 17:25





Ooops, a comma is missing. li $t4,105 The problem is that you can subtract an immediate from a register, but not subtract a register from an immediate in an instruction. So you put the immediate 105 in a register before the loop to perform the subtraction in an instruction.

– Alain Merigot
Mar 24 at 17:25













If you change sub $t2, $t1, $t2 # apply complement to $t2 to sub $t2, $t1, $t2 # apply complement to $t2 addi $t2, $t2, 48 all should work. --- nope. doesn't work.

– user366312
Mar 24 at 17:30





If you change sub $t2, $t1, $t2 # apply complement to $t2 to sub $t2, $t1, $t2 # apply complement to $t2 addi $t2, $t2, 48 all should work. --- nope. doesn't work.

– user366312
Mar 24 at 17:30













Onec corrected, your algorithm really seems correct, but I do not have time to test it. Maybe the problem comes from branches. Are your branches delayed ?

– Alain Merigot
Mar 24 at 17:37





Onec corrected, your algorithm really seems correct, but I do not have time to test it. Maybe the problem comes from branches. Are your branches delayed ?

– Alain Merigot
Mar 24 at 17:37













Are your branches delayed? --- nope.

– user366312
Mar 24 at 17:47






Are your branches delayed? --- nope.

– user366312
Mar 24 at 17:47




















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