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-2

I'm writing a program that converts hexadecimal values to the number digit value, to that end I've made it so the integer function htoi(char s[i]) accepts char arrays.

I plug in a char array using getchar() to allow the user to input characters which then get assigned to an array in my code, and said array is then used as the argument of htoi, however the compiler complains that htoi is using an argument with a pointer without a cast and wants char.

I've tried changing the arrays like char s[] to int s[i], because a char is just a smaller int, right? So, when I change it to int, the compiler then complains that htoi{s[i]) (s[] being an int now) makes a pointer from integer without a cast.

So, I realize that an iterator within an array is nothing but a pointer to an element within the array, okay. Great, so I remove I and. . . It expects an expression before ']' token.

I've exhausted my toolkit.

#include <stdio.h>
#define YES 1 
#define NO 0 
#define EXIT '~'

int htoi(int s[]); 

int main()
int i, c = 0;
int s[i];
for (i = 0; (c = getchar()!= EXIT); i++) c == ' ')
 htoi(s[]);
 else
 putchar(c);
 
 


int htoi(int s[])
int isHexidecimal;
int hexdigit; 
int n; 
int i = 0;


if(s[i] == '0')
 i++; 
 if(s[i] == 'x' 

isHexidecimal = YES;

while(isHexidecimal == YES)

if(s[i] >= '0' && s[i] <= 9)
 hexdigit += s[i] - '0';
 i++; 

else if(s[i] >= 'a' && s[i] <= 'f')
 hexdigit += s[i] - 'a' + 10; 
 i++; 

else if(s[i] >= 'A' && s[i] <= 'F')
 hexdigit += s[i] - 'A' + 10;
 i++;

else

 isHexidecimal = NO;

if(isHexidecimal == YES) 

 n = n * 16 + hexdigit;
 else 
 n += 0;
 
 
 return n;

I expect the main method to run the getchar() function which is assigned to variable c, and then assign variable c to the array, then pass said array onto htoi.

htoi would return the hexadecimal to integer value, or '0' if I have not entered a valid hexadecimal value, or returns the hexadecimal value and terminates as a result of the loop not having a valid hexadecimal value entered.

c

share|improve this question

asked Mar 26 at 22:48

Matthew_J_BarnesMatthew_J_Barnes

2377 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  int i, c = 0; int s[i]; Since i has no particular value, int s[i]; tries to declare an array with no particular size. Ouch.
  
  – David Schwartz
  Mar 26 at 23:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  OOF. cough Thanks, David-San. Alright, but I'm curious. Why does this trick work in Java but not in C? The chaining of assignments? I'm learning C through the C programming Language book 2nd edition, so some things are a little lost on me here.
  
  – Matthew_J_Barnes
  Mar 26 at 23:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Where is the chain in "i,c = 0" ? It's not "i = c = 0" as in stackoverflow.com/questions/13116833/… ... Only the "int" is kept with the , it doesn't try to assign to a tuple!
  
  – B. Go
  Mar 27 at 0:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So you need to set a max size to s, check it in the for loop (or risk a buffer overflow), and your current code won't work correctly for a second value input after one of your 3 separators. Actually as your string never ends with a , I'm not sure it will work even for the first value!
  
  – B. Go
  Mar 27 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Or any other valid stopping condition, as you're not using a string... Right now you parse uninitialized memory, it could contain anything after your value's chars
  
  – B. Go
  Mar 27 at 0:28
  

  
  
  
  

 | 
show 2 more comments

-2

I'm writing a program that converts hexadecimal values to the number digit value, to that end I've made it so the integer function htoi(char s[i]) accepts char arrays.

I plug in a char array using getchar() to allow the user to input characters which then get assigned to an array in my code, and said array is then used as the argument of htoi, however the compiler complains that htoi is using an argument with a pointer without a cast and wants char.

I've tried changing the arrays like char s[] to int s[i], because a char is just a smaller int, right? So, when I change it to int, the compiler then complains that htoi{s[i]) (s[] being an int now) makes a pointer from integer without a cast.

So, I realize that an iterator within an array is nothing but a pointer to an element within the array, okay. Great, so I remove I and. . . It expects an expression before ']' token.

I've exhausted my toolkit.

#include <stdio.h>
#define YES 1 
#define NO 0 
#define EXIT '~'

int htoi(int s[]); 

int main()
int i, c = 0;
int s[i];
for (i = 0; (c = getchar()!= EXIT); i++) c == ' ')
 htoi(s[]);
 else
 putchar(c);
 
 


int htoi(int s[])
int isHexidecimal;
int hexdigit; 
int n; 
int i = 0;


if(s[i] == '0')
 i++; 
 if(s[i] == 'x' 

isHexidecimal = YES;

while(isHexidecimal == YES)

if(s[i] >= '0' && s[i] <= 9)
 hexdigit += s[i] - '0';
 i++; 

else if(s[i] >= 'a' && s[i] <= 'f')
 hexdigit += s[i] - 'a' + 10; 
 i++; 

else if(s[i] >= 'A' && s[i] <= 'F')
 hexdigit += s[i] - 'A' + 10;
 i++;

else

 isHexidecimal = NO;

if(isHexidecimal == YES) 

 n = n * 16 + hexdigit;
 else 
 n += 0;
 
 
 return n;

I expect the main method to run the getchar() function which is assigned to variable c, and then assign variable c to the array, then pass said array onto htoi.

htoi would return the hexadecimal to integer value, or '0' if I have not entered a valid hexadecimal value, or returns the hexadecimal value and terminates as a result of the loop not having a valid hexadecimal value entered.

c

share|improve this question

asked Mar 26 at 22:48

Matthew_J_BarnesMatthew_J_Barnes

2377 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  int i, c = 0; int s[i]; Since i has no particular value, int s[i]; tries to declare an array with no particular size. Ouch.
  
  – David Schwartz
  Mar 26 at 23:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  OOF. cough Thanks, David-San. Alright, but I'm curious. Why does this trick work in Java but not in C? The chaining of assignments? I'm learning C through the C programming Language book 2nd edition, so some things are a little lost on me here.
  
  – Matthew_J_Barnes
  Mar 26 at 23:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Where is the chain in "i,c = 0" ? It's not "i = c = 0" as in stackoverflow.com/questions/13116833/… ... Only the "int" is kept with the , it doesn't try to assign to a tuple!
  
  – B. Go
  Mar 27 at 0:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So you need to set a max size to s, check it in the for loop (or risk a buffer overflow), and your current code won't work correctly for a second value input after one of your 3 separators. Actually as your string never ends with a , I'm not sure it will work even for the first value!
  
  – B. Go
  Mar 27 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Or any other valid stopping condition, as you're not using a string... Right now you parse uninitialized memory, it could contain anything after your value's chars
  
  – B. Go
  Mar 27 at 0:28
  

  
  
  
  

 | 
show 2 more comments

I'm writing a program that converts hexadecimal values to the number digit value, to that end I've made it so the integer function htoi(char s[i]) accepts char arrays.

I plug in a char array using getchar() to allow the user to input characters which then get assigned to an array in my code, and said array is then used as the argument of htoi, however the compiler complains that htoi is using an argument with a pointer without a cast and wants char.

I've tried changing the arrays like char s[] to int s[i], because a char is just a smaller int, right? So, when I change it to int, the compiler then complains that htoi{s[i]) (s[] being an int now) makes a pointer from integer without a cast.

So, I realize that an iterator within an array is nothing but a pointer to an element within the array, okay. Great, so I remove I and. . . It expects an expression before ']' token.

I've exhausted my toolkit.

#include <stdio.h>
#define YES 1 
#define NO 0 
#define EXIT '~'

int htoi(int s[]); 

int main()
int i, c = 0;
int s[i];
for (i = 0; (c = getchar()!= EXIT); i++) c == ' ')
 htoi(s[]);
 else
 putchar(c);
 
 


int htoi(int s[])
int isHexidecimal;
int hexdigit; 
int n; 
int i = 0;


if(s[i] == '0')
 i++; 
 if(s[i] == 'x' 

isHexidecimal = YES;

while(isHexidecimal == YES)

if(s[i] >= '0' && s[i] <= 9)
 hexdigit += s[i] - '0';
 i++; 

else if(s[i] >= 'a' && s[i] <= 'f')
 hexdigit += s[i] - 'a' + 10; 
 i++; 

else if(s[i] >= 'A' && s[i] <= 'F')
 hexdigit += s[i] - 'A' + 10;
 i++;

else

 isHexidecimal = NO;

if(isHexidecimal == YES) 

 n = n * 16 + hexdigit;
 else 
 n += 0;
 
 
 return n;

I expect the main method to run the getchar() function which is assigned to variable c, and then assign variable c to the array, then pass said array onto htoi.

htoi would return the hexadecimal to integer value, or '0' if I have not entered a valid hexadecimal value, or returns the hexadecimal value and terminates as a result of the loop not having a valid hexadecimal value entered.

c

share|improve this question

asked Mar 26 at 22:48

Matthew_J_BarnesMatthew_J_Barnes

2377 bronze badges

#include <stdio.h>
#define YES 1 
#define NO 0 
#define EXIT '~'

int htoi(int s[]); 

int main()
int i, c = 0;
int s[i];
for (i = 0; (c = getchar()!= EXIT); i++) c == ' ')
 htoi(s[]);
 else
 putchar(c);
 
 


int htoi(int s[])
int isHexidecimal;
int hexdigit; 
int n; 
int i = 0;


if(s[i] == '0')
 i++; 
 if(s[i] == 'x' 

isHexidecimal = YES;

while(isHexidecimal == YES)

if(s[i] >= '0' && s[i] <= 9)
 hexdigit += s[i] - '0';
 i++; 

else if(s[i] >= 'a' && s[i] <= 'f')
 hexdigit += s[i] - 'a' + 10; 
 i++; 

else if(s[i] >= 'A' && s[i] <= 'F')
 hexdigit += s[i] - 'A' + 10;
 i++;

else

 isHexidecimal = NO;

if(isHexidecimal == YES) 

 n = n * 16 + hexdigit;
 else 
 n += 0;
 
 
 return n;

share|improve this question

asked Mar 26 at 22:48

Matthew_J_BarnesMatthew_J_Barnes

2377 bronze badges

share|improve this question

asked Mar 26 at 22:48

Matthew_J_BarnesMatthew_J_Barnes

2377 bronze badges

asked Mar 26 at 22:48

Matthew_J_BarnesMatthew_J_Barnes

2377 bronze badges

asked Mar 26 at 22:48

Matthew_J_BarnesMatthew_J_Barnes

2377 bronze badges

1 Answer
1

active

oldest

votes

1

htoi(s[]);

is invalid, it doesn't mean anything. Just pass the pointer to the array.

htoi(s);

The

int htoi(int s[]);

The array declaration int s[] inside a function parameter list int htoi( <here> ) is interpreted exactly as if it is a pointer to the type int *s

share|improve this answer

edited Mar 26 at 23:04

answered Mar 26 at 22:52

Kamil CukKamil Cuk

18.9k22 gold badges88 silver badges3636 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  He may want to look at C11 Standard - 6.3.2.1 Other Operands - Lvalues, arrays, and function designators(p3) to figure out where the pointer you are talking about comes from.
  
  – David C. Rankin
  Mar 26 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I understand. Since I'm bringing up an array with an iterator, it works just like an array with a pointer, and therefore it doesn't work as I had intended.
  
  – Matthew_J_Barnes
  Mar 26 at 23:16
  

  
  
  
  

add a comment | 

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1

htoi(s[]);

is invalid, it doesn't mean anything. Just pass the pointer to the array.

htoi(s);

The

int htoi(int s[]);

The array declaration int s[] inside a function parameter list int htoi( <here> ) is interpreted exactly as if it is a pointer to the type int *s

share|improve this answer

edited Mar 26 at 23:04

answered Mar 26 at 22:52

Kamil CukKamil Cuk

18.9k22 gold badges88 silver badges3636 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  He may want to look at C11 Standard - 6.3.2.1 Other Operands - Lvalues, arrays, and function designators(p3) to figure out where the pointer you are talking about comes from.
  
  – David C. Rankin
  Mar 26 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I understand. Since I'm bringing up an array with an iterator, it works just like an array with a pointer, and therefore it doesn't work as I had intended.
  
  – Matthew_J_Barnes
  Mar 26 at 23:16
  

  
  
  
  

add a comment | 

1

htoi(s[]);

is invalid, it doesn't mean anything. Just pass the pointer to the array.

htoi(s);

The

int htoi(int s[]);

The array declaration int s[] inside a function parameter list int htoi( <here> ) is interpreted exactly as if it is a pointer to the type int *s

share|improve this answer

edited Mar 26 at 23:04

answered Mar 26 at 22:52

Kamil CukKamil Cuk

18.9k22 gold badges88 silver badges3636 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  He may want to look at C11 Standard - 6.3.2.1 Other Operands - Lvalues, arrays, and function designators(p3) to figure out where the pointer you are talking about comes from.
  
  – David C. Rankin
  Mar 26 at 22:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I understand. Since I'm bringing up an array with an iterator, it works just like an array with a pointer, and therefore it doesn't work as I had intended.
  
  – Matthew_J_Barnes
  Mar 26 at 23:16
  

  
  
  
  

add a comment | 

htoi(s[]);

is invalid, it doesn't mean anything. Just pass the pointer to the array.

htoi(s);

The

int htoi(int s[]);

The array declaration int s[] inside a function parameter list int htoi( <here> ) is interpreted exactly as if it is a pointer to the type int *s

share|improve this answer

edited Mar 26 at 23:04

answered Mar 26 at 22:52

Kamil CukKamil Cuk

18.9k22 gold badges88 silver badges3636 bronze badges

htoi(s[]);

htoi(s);

int htoi(int s[]);

share|improve this answer

edited Mar 26 at 23:04

answered Mar 26 at 22:52

Kamil CukKamil Cuk

18.9k22 gold badges88 silver badges3636 bronze badges

answered Mar 26 at 22:52

Kamil CukKamil Cuk

18.9k22 gold badges88 silver badges3636 bronze badges

answered Mar 26 at 22:52

Kamil CukKamil Cuk

18.9k22 gold badges88 silver badges3636 bronze badges