C Programming - Strings - Discussion

#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

If we assume that we are running this program on GCC compiler then for str1 and str2 pointers assigned strings will be 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.

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.

//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;
}

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).

It's all about memory nothing else.

str1 allocates 8 bytes.

But when we perform concate operation size of str1 extended. That's Why it gives error in old compilers.

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

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

I don't understand how you the output of program because the str3 gives us indiabix but the value of str1 is India. So how correct option is b? Please tell me.

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.

In the above program we are not using strcpy function to copy the concatenated value of s1 and s2 to s3. So I think it may result an error.