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printing characters stored in variables [duplicate]

Char (single letter) must be a pointer to correctly work?Is an array name a pointer?Arrays are Pointers?What does this guide mean by “And since arrays are actually pointers”?Incompatible integer to pointer conversion assigning to 'char *' from 'int'Why is my program not compiling?Getting warnings when assigning characters to a 2d array declared using malloc in CInitialising integer array as memberVigénere Cipher in C+ or - sign in array, CIn C89, I can't seem to make a character array from an existing oneWhile executing this program the char function does the returning the variable.i dont know why?Why do I get this error and warning? Error and warning is in the description

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0

This question already has an answer here:

• 
  Char (single letter) must be a pointer to correctly work?
  
  1 answer
  
  

I can't get my function to print characters i have initialised and declared.

#include <stdio.h>
int main() 

 char letter1 ="i";
 char letter2 ="n";
 char letter3 ="C";

 printf ("Programming %c%c %cn", letter1, letter2, letter3);


 return 0;

I want it to display

 "Programming in C"

using

 printf ("Programming %c%c %cn", letter1, letter2, letter3);

I get the following error

main.c:4:8: warning: incompatible pointer to integer conversion 
initializing 'char' with an expression of type 'char [2]' [-Wint- 
conversion]
 char letter1 ="i";
 ^ 

I haven't learn about pointers or anything yet, I'm used to simpler languages that just work. I am trying to go through an edX course but I find the quality to be poor and the pacing tedious.

Would be happy if you could help me out here and recommend me better resources for learning C

thanks

c

share|improve this question

asked Mar 24 at 20:21

Mr_BodgerMr_Bodger

141

marked as duplicate by melpomene, Daniel Pryden, Antti Haapala c
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  "i" is a string literal, you want character literals, which use a single quote 'i': en.cppreference.com/w/c/language/character_constant
  
  – UnholySheep
  Mar 24 at 20:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I didn't even know there was a difference, thanks it works now <3
  
  – Mr_Bodger
  Mar 24 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Or change char to char * and then do printf ("Programming %c%c %cn", *letter1, *letter2, *letter3); (otherwise character literals use single-quotes while string literals use double-quotes)
  
  – David C. Rankin
  Mar 24 at 20:47
  
  

  
  
  
  

add a comment | 

0

This question already has an answer here:

• 
  Char (single letter) must be a pointer to correctly work?
  
  1 answer
  
  

I can't get my function to print characters i have initialised and declared.

#include <stdio.h>
int main() 

 char letter1 ="i";
 char letter2 ="n";
 char letter3 ="C";

 printf ("Programming %c%c %cn", letter1, letter2, letter3);


 return 0;

I want it to display

 "Programming in C"

using

 printf ("Programming %c%c %cn", letter1, letter2, letter3);

I get the following error

main.c:4:8: warning: incompatible pointer to integer conversion 
initializing 'char' with an expression of type 'char [2]' [-Wint- 
conversion]
 char letter1 ="i";
 ^ 

I haven't learn about pointers or anything yet, I'm used to simpler languages that just work. I am trying to go through an edX course but I find the quality to be poor and the pacing tedious.

Would be happy if you could help me out here and recommend me better resources for learning C

thanks

c

share|improve this question

asked Mar 24 at 20:21

Mr_BodgerMr_Bodger

141

marked as duplicate by melpomene, Daniel Pryden, Antti Haapala c
Users with the  c badge can single-handedly close c questions as duplicates and reopen them as needed.

StackExchange.ready(function()
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$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
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$hover.hover(
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$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
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StackExchange.helpers.removeMessages();

);
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);
Mar 24 at 21:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  "i" is a string literal, you want character literals, which use a single quote 'i': en.cppreference.com/w/c/language/character_constant
  
  – UnholySheep
  Mar 24 at 20:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I didn't even know there was a difference, thanks it works now <3
  
  – Mr_Bodger
  Mar 24 at 20:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Or change char to char * and then do printf ("Programming %c%c %cn", *letter1, *letter2, *letter3); (otherwise character literals use single-quotes while string literals use double-quotes)
  
  – David C. Rankin
  Mar 24 at 20:47
  
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  Char (single letter) must be a pointer to correctly work?
  
  1 answer
  
  

I can't get my function to print characters i have initialised and declared.

#include <stdio.h>
int main() 

 char letter1 ="i";
 char letter2 ="n";
 char letter3 ="C";

 printf ("Programming %c%c %cn", letter1, letter2, letter3);


 return 0;

I want it to display

 "Programming in C"

using

 printf ("Programming %c%c %cn", letter1, letter2, letter3);

I get the following error

main.c:4:8: warning: incompatible pointer to integer conversion 
initializing 'char' with an expression of type 'char [2]' [-Wint- 
conversion]
 char letter1 ="i";
 ^ 

I haven't learn about pointers or anything yet, I'm used to simpler languages that just work. I am trying to go through an edX course but I find the quality to be poor and the pacing tedious.

Would be happy if you could help me out here and recommend me better resources for learning C

thanks

c

share|improve this question

asked Mar 24 at 20:21

Mr_BodgerMr_Bodger

141

#include <stdio.h>
int main() 

 char letter1 ="i";
 char letter2 ="n";
 char letter3 ="C";

 printf ("Programming %c%c %cn", letter1, letter2, letter3);


 return 0;

 "Programming in C"

 printf ("Programming %c%c %cn", letter1, letter2, letter3);

main.c:4:8: warning: incompatible pointer to integer conversion 
initializing 'char' with an expression of type 'char [2]' [-Wint- 
conversion]
 char letter1 ="i";
 ^ 

share|improve this question

asked Mar 24 at 20:21

Mr_BodgerMr_Bodger

141

share|improve this question

asked Mar 24 at 20:21

Mr_BodgerMr_Bodger

141

asked Mar 24 at 20:21

Mr_BodgerMr_Bodger

141

asked Mar 24 at 20:21

Mr_BodgerMr_Bodger

141

2 Answers
2

active

oldest

votes

0

 char letter1 ="i";

Unfortunately, "i" is a string. Use 'i' for characters or "i"[0] if you really want to.

If your compiler didn't give you a warning, it's either a terrible compiler or you don't have appropriate warnings turned on. If you got a warning and you ignored it ...

share|improve this answer

answered Mar 24 at 20:25

David SchwartzDavid Schwartz

141k14149237

add a comment | 

-2

Arrays in C are pointers to the first element. All other elements in the array follows the first element on stack stack. Strings in C are represented by char arrays. Therefore, following example will work:

char str[3];
str[0] = 'A';
str[1] = 'B';
str[2] = 'C';

Fortunately, C allows a shorthand for representing strings. Following are two examples:

char str[] = "ABC";
char str* = "ABC";

As previously mentioned, arrays are pointers. Therefore, both examples work. Also, note that we don't have enter a size of the array as we directly initialize the array.

If you really want to print a string only using char variables, then you have to use quotes, like the example below:

letter1 = 'A';
letter2 = 'B';
letter3 = 'C';

As literature, I highly recommend the book Problem Solving and Program Design in C by Hanly Koffman. It is simple to read and has a lot of code examples.

share|improve this answer

answered Mar 24 at 20:42

Martin PekárMartin Pekár

274

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Arrays are not pointers
  
  – Blastfurnace
  Mar 24 at 22:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  As I said, an array is a pointer to the first element.
  
  – Martin Pekár
  Mar 25 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Arrays are not, in fact, pointers. For example, given the definitions char arr[1], *ptr; it's a fact that sizeof(arr) != sizeof(ptr) and arr = ptr; is a syntax error. The array and the pointer are not the same types. I can find more questions/answers on Stack Overflow if you aren't convinced.
  
  – Blastfurnace
  Mar 25 at 8:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well, sizeof in C will return the amount of bytes for the data types times amount of element in the array. So, of course, they are not equal, since you have to get the address of an element in the array and compare its address size to the size of the pointer. And yes, ´arr=ptr´ is a syntax error, but the opposite ´ptr=arr´ is not. Technicallly, an array of a compile time specific size is allocated in the %rsp register, which is then pushed onto the stack. An array is a pointer to the first element in the %rsp register. To get the next element on the stack, add the byte size to the address.
  
  – Martin Pekár
  Mar 25 at 10:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also, why do you dynamically allocate memory in a pointer...?
  
  – Martin Pekár
  Mar 25 at 10:57
  

  
  
  
  

 | 
show 7 more comments

0

 char letter1 ="i";

Unfortunately, "i" is a string. Use 'i' for characters or "i"[0] if you really want to.

If your compiler didn't give you a warning, it's either a terrible compiler or you don't have appropriate warnings turned on. If you got a warning and you ignored it ...

share|improve this answer

answered Mar 24 at 20:25

David SchwartzDavid Schwartz

141k14149237

add a comment | 

0

 char letter1 ="i";

Unfortunately, "i" is a string. Use 'i' for characters or "i"[0] if you really want to.

If your compiler didn't give you a warning, it's either a terrible compiler or you don't have appropriate warnings turned on. If you got a warning and you ignored it ...

share|improve this answer

answered Mar 24 at 20:25

David SchwartzDavid Schwartz

141k14149237

add a comment | 

 char letter1 ="i";

Unfortunately, "i" is a string. Use 'i' for characters or "i"[0] if you really want to.

If your compiler didn't give you a warning, it's either a terrible compiler or you don't have appropriate warnings turned on. If you got a warning and you ignored it ...

share|improve this answer

answered Mar 24 at 20:25

David SchwartzDavid Schwartz

141k14149237

 char letter1 ="i";

share|improve this answer

answered Mar 24 at 20:25

David SchwartzDavid Schwartz

141k14149237

answered Mar 24 at 20:25

David SchwartzDavid Schwartz

141k14149237

answered Mar 24 at 20:25

David SchwartzDavid Schwartz

141k14149237

-2

Arrays in C are pointers to the first element. All other elements in the array follows the first element on stack stack. Strings in C are represented by char arrays. Therefore, following example will work:

char str[3];
str[0] = 'A';
str[1] = 'B';
str[2] = 'C';

Fortunately, C allows a shorthand for representing strings. Following are two examples:

char str[] = "ABC";
char str* = "ABC";

As previously mentioned, arrays are pointers. Therefore, both examples work. Also, note that we don't have enter a size of the array as we directly initialize the array.

If you really want to print a string only using char variables, then you have to use quotes, like the example below:

letter1 = 'A';
letter2 = 'B';
letter3 = 'C';

As literature, I highly recommend the book Problem Solving and Program Design in C by Hanly Koffman. It is simple to read and has a lot of code examples.

share|improve this answer

answered Mar 24 at 20:42

Martin PekárMartin Pekár

274

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Arrays are not pointers
  
  – Blastfurnace
  Mar 24 at 22:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  As I said, an array is a pointer to the first element.
  
  – Martin Pekár
  Mar 25 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Arrays are not, in fact, pointers. For example, given the definitions char arr[1], *ptr; it's a fact that sizeof(arr) != sizeof(ptr) and arr = ptr; is a syntax error. The array and the pointer are not the same types. I can find more questions/answers on Stack Overflow if you aren't convinced.
  
  – Blastfurnace
  Mar 25 at 8:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well, sizeof in C will return the amount of bytes for the data types times amount of element in the array. So, of course, they are not equal, since you have to get the address of an element in the array and compare its address size to the size of the pointer. And yes, ´arr=ptr´ is a syntax error, but the opposite ´ptr=arr´ is not. Technicallly, an array of a compile time specific size is allocated in the %rsp register, which is then pushed onto the stack. An array is a pointer to the first element in the %rsp register. To get the next element on the stack, add the byte size to the address.
  
  – Martin Pekár
  Mar 25 at 10:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also, why do you dynamically allocate memory in a pointer...?
  
  – Martin Pekár
  Mar 25 at 10:57
  

  
  
  
  

 | 
show 7 more comments

-2

Arrays in C are pointers to the first element. All other elements in the array follows the first element on stack stack. Strings in C are represented by char arrays. Therefore, following example will work:

char str[3];
str[0] = 'A';
str[1] = 'B';
str[2] = 'C';

Fortunately, C allows a shorthand for representing strings. Following are two examples:

char str[] = "ABC";
char str* = "ABC";

As previously mentioned, arrays are pointers. Therefore, both examples work. Also, note that we don't have enter a size of the array as we directly initialize the array.

If you really want to print a string only using char variables, then you have to use quotes, like the example below:

letter1 = 'A';
letter2 = 'B';
letter3 = 'C';

As literature, I highly recommend the book Problem Solving and Program Design in C by Hanly Koffman. It is simple to read and has a lot of code examples.

share|improve this answer

answered Mar 24 at 20:42

Martin PekárMartin Pekár

274

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Arrays are not pointers
  
  – Blastfurnace
  Mar 24 at 22:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  As I said, an array is a pointer to the first element.
  
  – Martin Pekár
  Mar 25 at 8:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Arrays are not, in fact, pointers. For example, given the definitions char arr[1], *ptr; it's a fact that sizeof(arr) != sizeof(ptr) and arr = ptr; is a syntax error. The array and the pointer are not the same types. I can find more questions/answers on Stack Overflow if you aren't convinced.
  
  – Blastfurnace
  Mar 25 at 8:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well, sizeof in C will return the amount of bytes for the data types times amount of element in the array. So, of course, they are not equal, since you have to get the address of an element in the array and compare its address size to the size of the pointer. And yes, ´arr=ptr´ is a syntax error, but the opposite ´ptr=arr´ is not. Technicallly, an array of a compile time specific size is allocated in the %rsp register, which is then pushed onto the stack. An array is a pointer to the first element in the %rsp register. To get the next element on the stack, add the byte size to the address.
  
  – Martin Pekár
  Mar 25 at 10:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Also, why do you dynamically allocate memory in a pointer...?
  
  – Martin Pekár
  Mar 25 at 10:57
  

  
  
  
  

 | 
show 7 more comments

Arrays in C are pointers to the first element. All other elements in the array follows the first element on stack stack. Strings in C are represented by char arrays. Therefore, following example will work:

char str[3];
str[0] = 'A';
str[1] = 'B';
str[2] = 'C';

Fortunately, C allows a shorthand for representing strings. Following are two examples:

char str[] = "ABC";
char str* = "ABC";

As previously mentioned, arrays are pointers. Therefore, both examples work. Also, note that we don't have enter a size of the array as we directly initialize the array.

If you really want to print a string only using char variables, then you have to use quotes, like the example below:

letter1 = 'A';
letter2 = 'B';
letter3 = 'C';

As literature, I highly recommend the book Problem Solving and Program Design in C by Hanly Koffman. It is simple to read and has a lot of code examples.

share|improve this answer

answered Mar 24 at 20:42

Martin PekárMartin Pekár

274

char str[3];
str[0] = 'A';
str[1] = 'B';
str[2] = 'C';

char str[] = "ABC";
char str* = "ABC";

letter1 = 'A';
letter2 = 'B';
letter3 = 'C';

share|improve this answer

answered Mar 24 at 20:42

Martin PekárMartin Pekár

274

answered Mar 24 at 20:42

Martin PekárMartin Pekár

274

answered Mar 24 at 20:42

Martin PekárMartin Pekár

274