C++ Q, num*num, resulting code

[StoneColdCrazy]

1).&nbsp;&nbsp;int a = 50*20;<br>2).&nbsp;&nbsp;int b=50, c=20, a=c*b;<br><br>Will the compiled code in case 1 multiply 20*50 in order to get the number a, or will that operation be performed at compile time??&nbsp;&nbsp;Same question about case 2??<br><br>NOTE: I using Borland C++, and the resulting code of case 1 from that compiler is, int a = 1000, (or something like that, except the code is in assembly), and the resulting code in case 2, is a = c*b...&nbsp;&nbsp;I just want to know, does all compilers do that, or some compilers will multipy the numbers at run time (even in 1st case)??

 

[Masi]

I think they all do the same thing while compiling. If they find a an '=' followed by a constant value ( can be also a constant result of an expression ) they optimize it.
So also if you have :
int a = 1+2+3+4+5;
the result is :
int a = 15;

In the second case, instead, <a> is a variable, so it's a bit more difficult for the optimizer to understand that you intend to use <a> as a constant. More intelligent optimizer do this and more, others not.