Code:
int a=5;
int b=a++ + ++a + a++;
printf("%d",b);
Code:
int a=5,b;
b=a++ + ++a + a++;
printf("%d",b);
why?
-
the output for this code is 19
but the output for this code is 18.
why? -
The C language standard provides no rules for when a compiler can change variable values inside a single statement. As a result, if you change the value of a variable more than once in a single statement, the compiler can make the changes in any order it chooses.
This is termed indeterminate results. -
Hi,
In my system both program give same answer 19,so may be it is depend on compiler.You use a for multiple time so,it is depend on compiler.There is no specific rules for compiler.Comment
-
I recommend this method over a for (t = 0; t < b; t++). Imagine if you were using 64-bit math, the for loop would take a long time, even on a computer. OP doubling method of a=a+a was on track, but incomplete.
unsigned a;
unsigned b;
unsigned product = 0;
scanf("%u%u", &a, &b);
while (b > 0) {
if (b & 1) {
unsigned OldProduct = product;
product += a;
if (product < OldProduct) {
printf("\nOverf low\n");
break;
}
b >>= 1; // Halve b
a += a; // Double a - avoid using multiplication
}
printf("\nProdu ct = %u\n", product);Comment
Comment