C Programming - Strings - Discussion
Discussion Forum : Strings - Find Output of Program (Q.No. 27)
27.
What will be the output of the program (Turbo C in 16 bit platform DOS) ?
#include<stdio.h>
#include<string.h>
int main()
{
char *str1 = "India";
char *str2 = "BIX";
char *str3;
str3 = strcat(str1, str2);
printf("%s %s\n", str3, str1);
return 0;
}
Answer: Option
Explanation:
It prints 'IndiaBIX IndiaBIX' in TurboC (in 16 bit platform).
It may cause a 'segmentation fault error' in GCC (32 bit platform).
Discussion:
19 comments Page 2 of 2.
Chatrapathi.s said:
1 decade ago
It leads to segmentation fault.
Keval Patel said:
1 decade ago
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.
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.
Razia said:
1 decade ago
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.
Prafull said:
10 years ago
*str3 does not initialize (or) does get 0 allocated memory. So it will give error.
Nani said:
9 years ago
If we use char so we need to use prints("%c") right there they using %s is it correct?
Karan said:
8 years ago
Please explain clearly.
Prashanth said:
7 years ago
Thanks @Vishal Kumar.
(1)
Pooja gurav said:
5 years ago
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.
(2)
Nitesh said:
5 years ago
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.
str1 allocates 8 bytes.
But when we perform concate operation size of str1 extended. That's Why it gives error in old compilers.
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