CODEWITHSUNDEEP
cstringscanf
P4Penguin
P4Penguin

Reputation: 23

Why does giving '&' for a string in scanf() give no error?

char text[10];
printf("Enter text[Max characters= 9]: ");
scanf("%[^\n]s",&text);

I wrote this by mistake and kept working with it, later on I noticed the mistake in scanf but my question is why was there no error? 3rd line means save the string at the address of address of first element of the array, correct?

Upvotes: 0

Views: 60

Answers (1)

Gabriel Pellegrino
Gabriel Pellegrino

Reputation: 1122

Your a.c source code

#include <stdio.h>
int main()
{
    char text[10];
    printf("%p\n", text);
    printf("%p\n", &text);
    printf("Enter text[Max characters= 9]: ");
    scanf("%[^\n]s",&text);
    printf("%p\n", text);
    printf("%p\n", &text);
    return 0;
}

As you can see on the following logs

bash-5.1$ gcc a.c -o a
bash-5.1$ ./a
0x7fff602b2166
0x7fff602b2166
Enter text[Max characters= 9]: afaefa
0x7fff602b2166
0x7fff602b2166

Text and &text points to the same address. &text on a scanf doesn't break your code but its redundant.

But if you compile with -Wall flag

bash-5.1$ gcc a.c -o a -Wall
a.c: In function ‘main’:
a.c:8:15: warning: format ‘%[^
   ’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[10]’ [-Wformat=]
    8 |     scanf("%[^\n]s",&text);
      |            ~~~^~    ~~~~~
      |               |     |
      |               |     char (*)[10]
      |               char *

Upvotes: 2

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