What will be the output of the program, if a short int is 2 bytes wide?
#include<stdio.h>
int main()
{
short int i = 0;
for(i<=5 && i>=-1; ++i; i>0)
printf("%u,", i);
return 0;
}
A.
1 ... 65535
B.
Expression syntax error
C.
No output
D.
0, 1, 2, 3, 4, 5
Answer: Option A
Explanation:
for(i<=5 && i>=-1; ++i; i>0)
so expression i<=5 && i>=-1 initializes for loop.
expression ++i is the loop condition.
expression i>0 is the increment expression.
In for( i <= 5 && i >= -1; ++i; i>0) expression i<=5 && i>=-1 evaluates to one.
Loop condition always get evaluated to true. Also at this point it increases i by one.
An increment_expression i>0 has no effect on value of i.so for loop get executed till the limit of integer (ie. 65535)
#include<stdio.h>
int main()
{
float a = 0.7;
if(0.7 > a)
printf("Hi\n");
else
printf("Hello\n");
return 0;
}
A.
Hi
B.
Hello
C.
Hi Hello
D.
None of above
Answer: Option A
Explanation:
if(0.7 > a) here a is a float variable and 0.7 is a double constant. The double constant 0.7 is greater than the float variable a. Hence the if condition is satisfied and it prints 'Hi' Example:
sizeof(str) denotes 6 * 4 bytes = 24 bytes. Hence it prints '24'
strlen(str[0])); becomes strlen(Frogs)). Hence it prints '5';
Hence the output of the program is 24, 5
Hint: If you run the above code in 16 bit platform (Turbo C under DOS) the output will be 12, 5. Because the pointer occupies only 2 bytes. If you run the above code in Linux (32 bit platform), the output will be 24, 5 (because the size of pointer is 4 bytes).
Input/output function prototypes and macros are defined in which header file?
A.
conio.h
B.
stdlib.h
C.
stdio.h
D.
dos.h
Answer: Option C
Explanation:
stdio.h, which stands for "standard input/output header", is the header in the C standard library that contains macro definitions, constants, and declarations of functions and types used for various standard input and output operations.
#include<stdio.h>
int main()
{
int i;
char c;
for(i=1; i<=5; i++)
{
scanf("%c", &c); /* given input is 'a' */
printf("%c", c);
ungetc(c, stdin);
}
return 0;
}
A.
aaaa
B.
aaaaa
C.
Garbage value.
D.
Error in ungetc statement.
Answer: Option B
Explanation:
for(i=1; i<=5; i++) Here the for loop runs 5 times.
Loop 1: scanf("%c", &c); Here we give 'a' as input.
printf("%c", c); prints the character 'a' which is given in the previous "scanf()" statement.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.
Loop 2:
Here the scanf("%c", &c); get the input from "stdin" because of "ungetc" function.
printf("%c", c); Now variable c = 'a'. So it prints the character 'a'.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.
This above process will be repeated in Loop 3, Loop 4, Loop 5.
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