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Login/Register#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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2 / 24
#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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3 / 24
#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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4 / 24
#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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5 / 24
#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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6 / 24
#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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7 / 24
#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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8 / 24
#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
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9 / 24
#include<stdio.h>
#include<string.h>
void main()
{
while (strcmp("some","some\0"))
printf("Strings are not equal\n");
}ANo Output No Error
BString are not equal
CCompilation Error
DNone of these
Answer: Option A
Explanation:Ending the string constant with \0 explicitly makes no difference.
So "some" and "some\0" are equivalent. So, strcmp returns 0 .
So breaking out of the while loop.
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10 / 24
#include<stdio.h>
void main()
{
char str1[] = {'s','o','m','e'};
char str2[] = {'s','o','m','e','\0'};
while (strcmp(str1,str2))
printf("Strings are not equal\n");
}ANo Output No Error
BString are not equal
Cstack overflow
DNone of these
Answer: Option C
Explanation:If a string constant is initialized explicitly with characters, '\0' is not appended automatically to the string.
Since str1 doesn't have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a '\0'. So str1 and str2 are not the same.
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