pointer printing

Why I cannot print the char d in the below code. I get segmentation fault :(

Code:

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#include <stdlib.h>
#include <stdio.h>

main(){
char a[]="Cool";
char b[]="things";
        char *c;
        c=a;
        char d;
        d=*(c+2);
       

        printf("%s", c);
        printf("\n %p", c);
        printf("\n %p", ++*c);
        printf("\n %p", *(c+2));
        printf("\n %p", d);
        printf("\n %s", d);
}

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this is the problem --> printf("\n %s", d); could you please help me how to print if that is possible?

d is a char. It is not a string, or a pointer, so interpreting it as such is a Bad Thing.

I think you forgot to null terminate your strings cool and things - when you use %s, it tries to print until it find '\0', which is not there, so it runs off the character array and segfaults.

If you want to just print the one character, use %c. If you want to print a string starting at that point, you need to terminate the string somehow.

Or not, that would just be my first guess at a place to begin to debug.

maybe make d an array like this

char d[100];

and to terminate string maybe use d*='/0'
pend a null character to end of string maybe.

im not sure about this though

What exactly are you trying to do? Print the address of d? that would be printf("Address of D is %p\n", &d); Print the value of d? That would be printf("Value of d is %d\n", (int) d);

If it is something else you intend, post it here. I am not surprised the code posted is segfaulting.

Quote:

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Originally Posted by KBriggs

I think you forgot to null terminate your strings cool and things - when you use %s, it tries to print until it find '\0', which is not there, so it runs off the character array and segfaults.

If you want to just print the one character, use %c. If you want to print a string starting at that point, you need to terminate the string somehow.

Or not, that would just be my first guess at a place to begin to debug.

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The string c is terminated, because a is terminated. The string b is also terminated. It would not be a source of problems if either string were involved. It works because the compiler will compute size requirements when you use the below syntax, and literal strings are always terminated.

char foo[] = "Cool and things";

jeffcobb is right. We need to know what c lady intends - the segfault is happening because she is using a format specifier, %s, with an incompatible value.

Really? I thought that declaring a character array like that without specifying a size would not null terminate it unless you do it explicitly.

I am going to go try it now ^_^


EDIT: you are right.

The segfault comes from trying to print a char (d) as a string (%s). Also, gcc really doesn't like the way the OP uses %p ^_^

Quote:

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Originally Posted by KBriggs

Really? I thought that declaring a character array like that without specifying a size would not null terminate it unless you do it explicitly.

I am going to go try it now ^_^


EDIT: you are right.

The segfault comes from trying to print a char (d) as a string (%s). Also, gcc really doesn't like the way the OP uses %p ^_^

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GCC apparently doesn't like the way most people do things around here...and I am glad. I know it is nowhere near all-powerful but it does catch a comforting amount of booboos.