Different output

[jimhakans]

Hello there. I need help with this program.

{ 
int a=10,b;
b=a++ + ++a;
printf("%d,%d,%d,%d",b,a++,a,++a);
}
 

The output of this program is 22,13,13,13. But in the printf statement "a++" should print 11 instead of 13, "a" should print 10 instead of 13 and "++a" should print 11 instead of 13. That is what i can make from this program. Can anybody explain me this code?

[Ganon11]

When you invoke the ++ operator (either pre- or post-increment) twice in the same statement, you invoke undefined behavior. How your compiler decides to resolve that is completely compiler-dependent. I'd be lucky to get the same output on my compiler.

[jimhakans]

Ok but if i reduces that program to this

{
int a=10;
printf("%d,%d",a,a++);
}
 

In this program my compiler is printing 11 first and then 10. Why?

[JosAH]

Ok but if i reduces that program to this

{
int a=10;
printf("%d,%d",a,a++);
}
 

In this program my compiler is printing 11 first and then 10. Why?

There's a paragraph in the C and C++ Standard that deals with "undefined behaviour".
Being dependent on the order of evaluation of your parameters simply induces
that undefined behaviour. Some compilers evaluate actual function parameters
from left to right; other evaluate them from right to left. You can't depend on it.

kind regards,

Jos

[jimhakans]

There's a paragraph in the C and C++ Standard that deals with "undefined behaviour".
Being dependent on the order of evaluation of your parameters simply induces
that undefined behaviour. Some compilers evaluate actual function parameters
from left to right; other evaluate them from right to left. You can't depend on it.

kind regards,

Jos

what if i want "a" to be print first, then what should i do?

[weaknessforcats]

what if i want "a" to be print first, then what should i do?

Use two printf() calls.

[kky2k]

When you invoke the ++ operator (either pre- or post-increment) twice in the same statement, you invoke undefined behavior. How your compiler decides to resolve that is completely compiler-dependent. I'd be lucky to get the same output on my compiler.

refer this you will get the whole idea....