dynamically allocated string populated with characters results in empty string C

Question

int main(){
    char *c=(char*)malloc(4*sizeof(char));
    *c='a';
    c++;
    *c='b';
    c++;
    *c='c';
    c++;
    *c='\0';
    printf("%s",c);
    return 0;
}

With this code I can print every single character (e.g. printf("%c",*(--c)); ), but when I try to print the whole string using printf("%s",c); the program prints nothing! Why is this happening ?

You have increased c to the last character '\0'... So nothing to print. — Jarod42
– Jarod42, Commented Aug 25, 2016 at 9:56
That really was an error you should have been able to detect yourself (yes, that includes beginners). Next time please use a debugger, single step and watch the variables before asking. — too honest for this site
– too honest for this site, Commented Aug 25, 2016 at 11:38

Jay · Accepted Answer · 2016-08-25 10:46:13Z

6

The pointer has been moved. So it is now pointing to the last character which is a "\0". So, nothing is printed. Preserve the pointer in another variable and try to print using that. It will then print correctly.

Try like below:

int main()
{
    char *p;
    char *c=(char*)malloc(4*sizeof(char));
    p = c;
    *c='a';
    c++;
    *c='b';
    c++;
    *c='c';
    c++;
    *c='\0';
    printf("%s",p);
    free(p);
    return 0;
}

edited Aug 25, 2016 at 10:46

answered Aug 25, 2016 at 9:58

Jay

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Comments

Jarod42 · Accepted Answer · 2016-08-25 10:00:44Z

1

You have increased c to the last character '\0'... So nothing to print.

You may change your code to:

int main(){
    char *p = (char*)malloc(4 * sizeof(char));
    char* c = p;
    *c = 'a';
    c++;
    *c = 'b';
    c++;
    *c = 'c';
    c++;
    *c = '\0';
    printf("%s", p);
    free(p);
    return 0;
}

or

int main(){
    char *c = (char*)malloc(4 * sizeof(char));

    c[0] = 'a';
    c[1] = 'b';
    c[2] = 'c';
    c[3] = '\0';
    printf("%s", c);
    free(c);
    return 0;
}

answered Aug 25, 2016 at 10:00

Jarod42

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Comments

KostasRim · Accepted Answer · 2016-08-25 10:01:07Z

1

c is a pointer and therefore the expression c++; changes the pointer location. After calling c++ 3 times you end up with c pointing to \0. To fix this issue create a temporary pointer to iterate through the elements. Also you program has a memory leak. Do not forget to free the memory!

edited Aug 25, 2016 at 10:01

answered Aug 25, 2016 at 9:58

KostasRim

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Comments

someone_ smiley · Accepted Answer · 2016-08-25 10:04:31Z

1

Because your string is now pointed to last character in allocated string. Try something like

int main(){
    char *c=(char*)malloc(4*sizeof(char));
    char* temp = c;
    *c='a';
    c++;
    *c='b';
    c++;
    *c='c';
    c++;
    *c='\0';
    printf("%s",c); // as a result of ++ operator, your pointer is no longer point at the beginning of string
    printf("%s,temp"); // << This will print whole string as it still pointed to the first address of the string
    free(temp);
    return 0;
}

edited Aug 25, 2016 at 10:04

answered Aug 25, 2016 at 9:59

someone_ smiley

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1 Comment

user2697817 Over a year ago

Where is the memory leak?

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