C Programming - Strings - Discussion

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.

Thanks @Vishal Kumar.

#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

//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;
}

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.

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.

Please explain clearly.

If we use char so we need to use prints("%c") right there they using %s is it correct?

*str3 does not initialize (or) does get 0 allocated memory. So it will give error.

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.