Question

I am refreshing my C skills. I am using a char *s and using malloc to allocate memory to the s. Then using scanf, I read the input to s. But my question is I haven’t specified a size for the memory chunk. But the program works. How does the memory gets allocated for the arbitrary length of the input string? Is scanf simply incrementing the pointer and writing data into the location?

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

int main() {
    char *s;
    s = (char *) malloc(sizeof(s));    //I did not specify how much like malloc(sizeof(s) * 128)
    if (s == NULL) {
        fprintf(stderr, "\nError allocating memory for string");
        exit(1);
    }
    scanf("%s", s);
    puts(s);
    free(s);
    return 0;
}

/*
    Input:
    aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
    Output:
    aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
*/

Answer 1

With char *s;, sizeof(s) is the same as sizeof(char *) which is either 4 or 8 depending on whether you are on a 32 bit box or a 64 bit box.

IF you are on a 32 bit box then you can store 3 characters plus the null ‘end of string’ character. IF you store more it may explode.