C Programming - Strings - Discussion

If we assume that string literals (e.g. "India", "BIX") are stored in code segment of program, then all given options of answers are incorrect. The correct option should have been run-time error or segmentation fault. Above program is trying to modify string stored in code (read-only) segment of program. Online compiler gives segmentation fault.

There is no memory allocated to str3. Hence it points to null.

So the statement str3 = strcat(str1, str2); will cause a segmentation fault.

There are 2 results after the strcat line:
- str1 = str1 concatenate str2 <- segmentation fault. str1 has only 6 byte in memory.
- str3 = str1 <- str3 will point to the same memory as str1 do.

If we write char str1[8] instead of char *str1, then the code will work (tested in GCC).

In 32 it causes segmentation faut.

If I take the example 32bits gcc compiler, the answer should be D.

It will give segmentation fault as we are trying to write read only memory.

As we write char *str3; it is stored in Read only memory (code segment) in GCC.

So we can not change the content. So it gives Segmentation fault.

#include <stdio.h>

char *strcat(char *dest, const char *src);

Description:

The strcat function concatenates or appends src to dest. All characters from src are copied including the terminating null character.

Return Value:

The strcat function returns dest.

Example:

#include <stdio.h>

int main() {
char string1[20];
char string2[20];

strcpy(string1, "Hello");
strcpy(string2, "Hellooo");

printf("Returned String : %s\n", strcat( string1, string2 ));
printf("Concatenated String : %s\n", string1 );

return 0;
}

It will proiduce following result:

Returned String : HelloHellooo
Concatenated String : HelloHellooo

Sir in printf statement have str1 is India but the result is indiabix. How?

//This works, problem was str1 was not having enough memory.

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

int main()
{
char str1[10] = "India";
char *str2 = "BIX";
char *str3;
str3=strcat(str1, str2);
printf("%s\n%s", str1,str3);
return 0;
}