GATE Overflow | Programming | Test 1 | Question: 22

Question

Which one among the following definitions of string str could cause problem when passed as the first argument to printf function?

• A. char str[] = "Hello World";
• B. char str[12] = "Hello World";
• C. char *str = "Hello World";
• D. char str[] = {'H','e','l','l','o',' ','W','o','r','l','d'};

Comments

M sorry if it sounds trivial to u bt y D) will cause problem?

I compiled and run it and it worked fine

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char str[] = {'H','e','l','l','o',' ','W','o','r','l','d', '\0'};

Answer 1

We are asked,

Which one among the following definitions of string str could cause a problem when passed as the first argument to printf function?

Option D. char str[] = {'H','e','l','l','o',' ','W','o','r','l','d'};   // It is just a character array but not string. 

Now if we add a null character '\0' at the end of all characters in the above array then it will be string str= Hello World.

And in the rest options, the compiler can automatically append a null character (\0) to the end of the array and treat the character array as a character string.

Comments

In option D, compiler just warn but giving the same output.

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read full quetion. you have to print the full array . printf("%s", str); give gabage value + end is not possible some time.

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ni am not getting any problem so why option d is wrong?

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"Could cause a problem" means there is a chance of error. Every string in C must end with a '\0', i.e., 8 bits of 0 or else printf and other functions expecting a string will keep on going till it finds one such '\0'; If no such byte exists until allocated process memory, segmentation fault or similar memory error comes.

In C language, just because a code runs on a system perfectly, does not guarantee that the code is correct.

Answer 2

answer (d) is correct becuse d is a character array not a string but all 3 are string declration

Answer 3

In C a string is defined as an array of characters termined by the NULL terminator, ie, \0.

In Options A, B and C, the compiler would implicitly add a NULL terminator at the end.

In Option D, we're defining the whole character array ourselves and we didn't add the NULL terminator explicitly, which is required here.

Answer 4

#include <stdio.h>
int main(){
    char Str1[] = "Hello World";
    char Str2[12] = "Hello World";
    char *Str3 = "Hello World";
    char Str4[] = {'H','e','l','l','o',' ','W','o','r','l','d'};
    printf("%s\n",Str4);
    printf("Sizeof Str1 : %d \t Strlen Str1 : %d\n",sizeof(Str1),strlen(Str1));
    printf("Sizeof Str2 : %d \t Strlen Str2 : %d\n",sizeof(Str2),strlen(Str2));
    printf("Sizeof Str3 : %d \t Strlen Str3 : %d\n",sizeof(Str3),strlen(Str3));
    printf("Sizeof Str4 : %d \t Strlen Str4 : %d\n",sizeof(Str4),strlen(Str4));
}

Output : 

Hello World
Sizeof Str1 : 12 	 Strlen Str1 : 11
Sizeof Str2 : 12 	 Strlen Str2 : 11
Sizeof Str3 : 8 	 Strlen Str3 : 11  
Sizeof Str4 : 11 	 Strlen Str4 : 11  

You will Not get any Error that's correct But, If you see the String4 sizeof(Str4).  i.e 11 which means the compiler is not adding the '\0' in Str4.

Sizeof (Str3) is 8 because its pointer type variable (i.e Sizeof(pointer) = 8 Byte

For all remaining String from Str1, Str2, Str3 the compiler Implicit added the '\0'     

Comments

If compiler is not adding \0 character after ending of option D, then how strlen() function terminates & gives the correct result.