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C Programming :: String

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Predict the output or error(s) for the following:

void main()
{
char string[]=Hello World;
display (string);
}

void display(char *string){
printf(%s,string);
}


ACompiler Error

BHello World

CHello

DWorld

Answer: Option A

Explanation:

Here is no explanation for this answer

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char a[5] ="hello"
What is wrong in the above syntax?


AIn array we can't do the operation

BSize of a is too large

CSize of a is too small

DNothing wrong with it

Answer: Option C

Explanation:

Here is no explanation for this answer

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What is the output of the following problem ?
#include<stdio.h>
int main()
{
char *c;
c = "Hello";
printf("%s\n", c);
return 0;
}

AHello

BH

Cello

DCompilation Error

Answer: Option A

Explanation:

Here is no explanation for this answer

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What is the output of the following problem ?
#include<stdio.h>
int main()
{
char *str = "12345";
printf("%c %c %c\n", *str, *(str++),  *(str++));
return 0;
}

A3 2 1

B1 2 3

C3 4 5

DCompilation Error

ENone of these

Answer: Option A

Explanation:

Here is no explanation for this answer

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What is the output of the following problem ?
#include<stdio.h>
int main()
{
int len=4;
char *st="12345678"; 
st = st + len; 
printf("%c\n",*st);
return 0;
}

A5

B4

C12345678

DCompilation Error

ESegmentation fault

Answer: Option A

Explanation:

Here is no explanation for this answer

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What is the output of the following problem ?
#include<stdio.h>int main()
{
char string[]="Hello World";
display(string);
return 0;
}
void display(char *string)
{
printf("%s",string);
}

AHello World

BHello

CWorld

DCompiler Error

ENone of these

Answer: Option E

Explanation:

In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

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What is the output of the following problem ?
#include<stdio.h>
void  main() 
{
char a[]="12345\0";
int i=strlen(a);
printf("Value of i %d\n",++i);
}

AValue of i 6

BValue of i 5

CValue of i 7

DCompilation Error

ENone of these

Answer: Option A

Explanation:

The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, but after the pre- increment in the printf statement, value of I will be printed.

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What is the output of the following problem ?
#include<stdio.h>
void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok Done \n"); 
else
printf("Forget it\n"); 
}

AForget it

BOk Done

CCompilation Error

DNone of these

Answer: Option B

Explanation:

Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok
Done" will be printed.

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What is the output of the following 'C' program ?
#include<stdio.h>
void main()
{
char str1[] = "Hello";
char str2[] = "Hello";
if (str1 == str2)
printf("\nequal");
else
printf("\nUnequal");
}

AEqual

BError

CUnequal

DNone of these

Answer: Option C

Explanation:

The '==' operator is meant for reference comparison i.e, it returns true only if the 2 string objects are same, else it will return false. Here, since the 2 string objects are different, the else part gets executed, and the output is Unequal, i.e, option C. 


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What is the output of the following C Program?
#include<stdio.h>
#include<string.h>
void main() {
int i, n;
char *x = "girl";
n = strlen(x);
*x = x[n];
for(i=0; i<n; ++i)
{
printf("%s\n",x);
x++;
}}

ASegmentation fault

Bgirl irl rl l

C(blank space) irl rl l

DCompilation Error

Answer: Option A

Explanation:

Here is no explanation for this answer

Workspace

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