CODEWITHSUNDEEP
cgcc
Veske
Veske

Reputation: 557

C - Strings are giving me errors

So I am writing a simple code to print out each symbol in my string. When compiling it, it gives me an error tough that I do not understand:

The code:

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

int main (void) {
    char my_string[50];
    int i, n;
    printf("Type in a string please : ");
    scanf("%s", &my_string);
    n = strlen(my_string);
    for (i = 0;i < n; i++) {
        printf("%c",my_string[i]);
    }

}

The error it gives:

gcc yl2.c -o Yl2
yl2.c: In function ‘main’:
yl2.c:9:2: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[50]’ [-Wformat=]
  scanf("%s", &my_string);
  ^

What is the problem here?

Upvotes: 1

Views: 1281

Answers (4)

Hafiz Muhammad Shafiq
Hafiz Muhammad Shafiq

Reputation: 8670

remove & sign in scan, it will work Also no need to use string.h here

Upvotes: 1

doctorlove
doctorlove

Reputation: 19232

&my_string is a pointer to my_string i.e. a ‘char (*)[50]’

You can either use 

scanf("%s", &my_string[0]);

or more conventionally

 scanf("%s", my_string);

Upvotes: 2

Lundin
Lundin

Reputation: 213276

scanf("%s", &my_string); should be scanf("%s", my_string);

my_string is an array, which decays to a pointer when you type it out without [ ]. If you type &my_string, you get an array pointer, which is strictly speaking not the same thing. That's why the compiler complains.

Upvotes: 2

Sunil Bojanapally
Sunil Bojanapally

Reputation: 12658

scanf("%s", &my_string); there should not be a &.

Since you have declared char my_string[50]; my_string as character array which is of type char * which is expecting in scanf() as the warning states.

Just use, scanf("%s", my_string);. Base address of the array as argument is sufficient.

Upvotes: 6

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