Can some one explain why the output of program is
0 1 1 3 1
void main(void)
{
int i=-1,j=0,k=1,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Main concern is "why k is not incremented".
FYI..I am compiling the program in VC++ editor Windows 7 32 bit. Many Thanks in advance.
Roughly:
To evaluate i++&&j++, compiler evaluated i first. The result is -1. -1 is stored in a temporary variable. Then i got incremented.
Because -1 is not zero, compiler evaluated j, which is 0. Compiler now evaluated -1 && 0, which is 0. Then j got incremented.
At this point, i = 0 and j = 1. Remaining expression: m=0&&k++||l++;
To evaluate 0&&k++, compiler noted that the first operand is 0. The result must be 0 so compiler didn't evaluate k or k++. Remaining expression: m=0||l++;
I hope you can do the rest. :)
Lets break this down into its separate operations:
i++ && j++: This will be the same as -1 && 0, which is false (i.e. 0).i and j are then increased to 0 and 1 respectively.0 && k++: The zero is from the previous logical operation, the result is false since the first operator is false.k is not increased, because of the shortcut nature of the logical operators.0 || l: The zero is still from the previous logic operation, it is 0 || 2 and the result will be true, i.e. 1.l is increased to 3.m, which now becomes true (i.e. 1)The whole expression causes i, j and l to be increased, and m to become 1. Just the result you are seeing.
0 && 1, the result is guaranteed to be 0, and so the right hand side of the && operator needs not be evaluated. Same with 1 || 0, it will always be 1 and the same thing happens. This is all specified in the C specification.
your value is getting calculated like below
m=((((i++)&&j++)&&k++)||l++);
since all ++ are post increment so during calculation of m all variable have same value what you have initialised but on the next line during print they all are incremented.Last is || so final TRUE will return to value of m.
void main()