A beginner level problem with char datatype

This is a discussion on A beginner level problem with char datatype within the C Programming forums, part of the General Programming Boards category; Hi all, I am a new user of c language and I have written a program which just first stores ...

  1. #1
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    A beginner level problem with char datatype

    Hi all,
    I am a new user of c language and I have written a program which just first stores the char values in memory then it reads and prints. Following is the code:

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    #include <math.h>
    int main()
    {
    	
    	int n1,i;
    	int buff_size = 26;
    	char * y;
    
    	y = (char *)malloc(sizeof(char)*buff_size);
    	if (y == NULL) {
    	fprintf(stderr, "Malloc returned void pointer\n");
    	exit(EXIT_FAILURE);
        }
    	for (i=0; i<buff_size; i++){
    		y[i] = (char)(i);
    		printf('value stored  is : %c \n' ,  y[i]);
    	}	
    
        return 0;
    }
    This code is not compiling giving the following error message:
    Code:
    error C2015: too many characters in constant  (at the following  line) 
    
    printf('value stored  is : %c \n' ,  y[i]);
    what could be the problem??

    Thanks

  2. #2
    spurious conceit MK27's Avatar
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    Code:
    		printf('value stored  is : %c \n' ,  y[i]);
    Those should be "double quotes", not 'single quotes'.

    You may also want to replace %c with %d since 0-26 are not printable characters and may screw up your console -- but try it both ways and see.
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  3. #3
    Programming Wraith GReaper's Avatar
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    Code:
    printf("value stored  is : %c \n" ,  y[i]);
    Devoted my life to programming...

  4. #4
    Registered User hk_mp5kpdw's Avatar
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    Code:
    y = (char *)malloc(sizeof(char)*buff_size);
    When using C, avoid typecasting the return value of the malloc call. Also, sizeof(char) is guaranteed to always be 1, thus the above could be safely reduced to:
    Code:
    y = malloc(buff_size);
    You should also free the memory you've allocated to y at the end of your program (before the return statement):
    Code:
    free(y);
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
    -Christopher Hitchens

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