c++ - why does left shift with variables generate different result from that with constant? -


after compiling code below, got weird result, =1 while b =0. explain what's going on behind scene?

#include<iostream> using namespace std;  int main(){      int n=32;      int a=1<<n;  //turns out a=1     int b=1<<32; //turns out b=0     cout<<"a="<<a<<endl;     cout<<"b="<<b<<endl; }

the standard not define, or rather, defines "undefined behaviour", happens in case left shift beyond size of integer type. [the reason undefined behaviour different hardware may or may not behave same given, example, 32-bit shift left].

in first case [at least without optimization], compiler generates instructions calculate 1 << 32 - on x86 turns 1 << (32 & 31) same 1 << 0 - 1.

in second case, compiler calculate value itself, turns overflow, , gives zero.

it's quite (but not certain) if change compiler options optimize code, gives 0 both cases. , going behaviour want if loop of smaller shifts (although may find "intersting" behaviour fact number turns negative, best use unsigned shift operations, really).


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