用scanf和C中的字符串进行分段错误

Question

I am a beginner with c, and I am having a problem with scanf and strings.

here is an example I wrote of my problem.

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

int main(void)
{
    char* string;
    scanf("%s", &string);
    if (strcmp(string, "Foo") == 0)  //segmentation fault here
        printf("Bar");
}

basically, this code compiles, but when I run it I get a segmentation fault in strcmp()

if I replace the "string" in that line with "&string" it works, but I get this error from the compiler

/usr/include/stdio.h:362:12: note: expected 'const char * __restrict__' but argument is of type 'char **'

which makes me think that this solution is not really ideal.

also If I declare string like this:

char string[100];

that works without any warnings, but that is also not ideal because I am not sure how large the string is going to be.

Is there a better solution I'm missing here, or are these my only options?

thank you.

Answer 1

char* string;
scanf("%s", &string);

string is not pointing to any valid memory location. Allocate memory using malloc to an array of characters and copy input to it. Make sure allocated memory has space for null termination character. Remember to free the memory to avoid leaks.

Answer 2

Just try that code

     #include <stdio.h>
     #include <string.h>
     #include <stdlib.h>
     int main(void)
     {
       char* string;
       string=(char *)malloc(3); /*allocate the memory to string cahr pointer(default pointer point to single byte and if you print pointer variable don't used & character)*/                                                      
       scanf("%s", string);
       if (strcmp(string, "Foo") == 0)  
            printf("Bar\n");
     }

Answer 3

While declaring a char *. it will not having any memory location. so you have to allocate a memory location before you use that variable. char *p; p=malloc(sieof(char) * size of string);

then you use scanf() function. it will work properly.

when we are accessing a unknown memory(ie unallocated memory). then it will through the segmentation fault