Determine Output:
void main()
{
int i=-1, j=-1, k=0, l=2, m;
m = i++ && j++ && k++ || l++;
printf("%d %d %d %d %d", i, j, k, l, m);
}
void main()
{
int i=-1, j=-1, k=0, l=2, m;
m = i++ && j++ && k++ || l++;
printf("%d %d %d %d %d", i, j, k, l, m);
}A. 0 0 1 2 0
B. 0 0 1 3 0
C. 0 0 1 3 1
D. 0 0 0 2 1
Answer: Option C
Solution(By Examveda Team)
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression 'i++ && j++ && k++' is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for '0 || 0' combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.
This Question Belongs to C Program >> C Miscellaneous
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Comments ( 1 )
Determine output:
void main()
{
int const *p=5;
printf("%d", ++(*p));
}
A. 6
B. 5
C. Garbage Value
D. Compiler Error
A. mmm nnn aaa
B. mmmm nnnn aaaa
C. Compiler Error
D. None of These
A. I hate Examveda
B. I love Examveda
C. Error
D. None of These
Determine Output:
void main()
{
static int var = 5;
printf("%d ", var--);
if(var)
main();
}
A. 5 5 5 5 5
B. 5 4 3 2 1
C. Infinite Loop
D. None of These
main()
{ int i = 0 , j = 1 , k =2 , l ;
l = i && j++ && ++k ;
printf ( “ %d %d %d %d ” , i , j , k , l ) ;
}
Ans: Output : 0 1 2 0
why out put is not : 0 2 3 0