CODEWITHSUNDEEP
cpointerssegmentation-fault
tunetopj
tunetopj

Reputation: 581

why does c allow initialization of string without declaration?

When the arguments of dyn_mat are constants, the code runs through without any error and s1 and s2 do store the input values.

#include<stdio.h>

int main(int argc, char const *argv[])
{
    char *s1, *s2;
    int n1=7, n2=8;
    printf("Enter, %d \n", n1);

    scanf("%s", s1);
    scanf("%s", s2);

    int dyn_mat[155][347];

    return 0;
}

but with arguments as variables, say n1 and n2, scanf reading s1 gives segmentation fault.

Upvotes: 1

Views: 129

Answers (2)

alk
alk

Reputation: 70971

why does c allow initialization of string without declaration?

There is no data type string in C.

In C one possible way to store a string of characters is using an array of characters, with the last element of this array carring a 0 to indicate the end of this string.

You program does not declare any array, but just pointers to characters, which have no memory assigned to which you copy data using scanf().

Your just lucky the program does not crash with the first call to scanf().

Upvotes: 0

Kerrek SB
Kerrek SB

Reputation: 477444

The code simply has undefined behaviour, since s1 and s2 are not valid pointers. scanf expects a pointer to an array of chars that's large enough to hold the read data, and you are not providing such pointers.

The usual way would be something like this:

char s1[1000];
char s2[1000];

scanf("%s", s1);
scanf("%s", s2);

(Though you should use a safer version that specifies the available buffer size rather than hoping for the input to be sufficiently short; for example, scanf("%999s", s1);.)

Upvotes: 7

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