int main()
{
int a, b, c;
a = 10;
b = 20;
c = printf("%d", a) + ++b;
printf("\n%d", c);
}
The output of the above program is 23 it seems but i dont know how it is obtained. Can anyone have an idea about it?
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3 Answers
printf has a return value, which is the total number of characters it prints.
The statement printf("%d",a) will print 10, which means the return value of printf here is 2.
The rest is easy:
c=printf("%d",a)+ ++b;
c will have a value of 2 + 20 + 1, which is 23.
3 Comments
dcaswell
Excellent explanation. It's also worth pointing out that b is 21 at the end of this function. It got incremented on this line:
c=printf("%d",a)+ ++b;
Yu Hao
@user814064 Right, that's the side effect. Your answer is correct but it seems that you have deleted it.
dcaswell
You were first by a few seconds -- instead I voted up your post.
Here the output will be two different integers,for two different printf statements . For the first printf statement the code prints 10, then when this printf statement participate in some assignment statement , it is treated as the number of characters it is printing i.e. 2 here. Then it is added to ++b i.e. 21 (PRE-INCREMENTED) . So the output is 23(2 + 21) .
The whole output looks like this :
10
23
Comments
printf returns the number of characters printed as an integer. So as you are printing 10 it will return 2. So now
c=printf("%d",a)+ ++b; will become
c=2+ ++b;
since b with a value of 20 is pre-incremented this will become
c=2+21 Therefore c=23
