Read string C - Linux

Question

I'm trying to read a string and print it in Linux using:

cc child.c -o child

#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[]) {  

    char st[100];

    scanf("%s", &st);
    printf("%s", st);

    return 0;
}

However I encounter this warning that will not allow me to compile.

child.c: In function ‘main’:
child.c:8:2: warning: format ‘%s’ expects argument of type ‘char ’, but argument 2 has type ‘char ()[100]’ [-Wformat=] scanf("%s", &st);

How can I solve it? Thanks!

Works fine for me! (Only think to change: scanf("%s", &st); to scanf("%s", st);) — Rizier123
– Rizier123, Commented Dec 17, 2014 at 22:41

haccks · Accepted Answer · 2014-12-17 22:42:23Z

4

In C, when passed to a function, array names converted to pointer to the first element of array. No need to place & before st. Change

scanf("%s", &st);

to

scanf("%s", st);

answered Dec 17, 2014 at 22:42

haccks

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Comments

datenwolf · Accepted Answer · 2014-12-17 22:41:49Z

2

In C array types degenerate to pointers. So when you take the address of an array, you actually get a pointer to a pointer. Just pass the array itself to scanf.

answered Dec 17, 2014 at 22:41

datenwolf

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