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1 / 24
What is the output of the following problem ?
#include
int main()
{
char *c;
c = "Hello";
printf("%s\n", c);
return 0;
}
Answer: Option A
Explanation:Here is no explanation for this answer
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2 / 24
What is the output of the following problem ?
#include
int main()
{
char *str = "12345";
printf("%c %c %c\n", *str, *(str++), *(str++));
return 0;
}
Answer: Option A
Explanation:Here is no explanation for this answer
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3 / 24
What is the output of the following problem ?
#include
int main()
{
int len=4;
char *st="12345678";
st = st + len;
printf("%c\n",*st);
return 0;
}
Answer: Option A
Explanation:Here is no explanation for this answer
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4 / 24
int main()
{
char string[]="Hello World";
display(string);
return 0;
}
void display(char *string)
{
printf("%s",string);
}
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
void main()
{
char a[]="12345\0";
int i=strlen(a);
printf("Value of i %d\n",++i);
}
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
void main()
{
int i;
char a[]="\0";
if(printf("%s\n",a))
printf("Ok Done \n");
else
printf("Forget it\n");
}
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
What is the output of the following 'C' program ?
#include
void main()
{
char str1[] = "Hello";
char str2[] = "Hello";
if (str1 == str2)
printf("\nequal");
else
printf("\nUnequal");
}
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
What is the output of the following C Program?
#include
#include
void main() {
int i, n;
char *x = "girl";
n = strlen(x);
*x = x[n];
for(i=0; i
printf("%s\n",x);
x++;
}}
Answer: Option A
Explanation:Here is no explanation for this answer
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9 / 24
#include
#include
void main()
{
while (strcmp("some","some\0"))
printf("Strings are not equal\n");
}
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
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");
}
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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