Question

我想知道，当两个整数相乘并且结果是类型转换为short并指定为short时，编译器会将它解析为什么？以下是代码段

int a=1,b=2,c;
short x=3,y=4,z;

int p;
short q;

int main()
{

c = a*b; /* Mul two ints and assign to int
            [compiler resolves this to __mulsi3()] */

z = x*y; /* Mul two short and assign to short
            [compiler resolves this to __mulhi3()] */

p = (x*y); /* Mul two short and assign to int
              [compiler resolves this to __mulsi3()] */

q =(short)(a*b); /* Mul two ints typecast to short and assign to short
                    [compiler resolves this to __mulhi3()] */

return 0;

}

在q =(short)(a*b);的情况下，应首先进行两次整数乘法（使用__mulsi3()），然后将其指定为short。但在这种情况并非如此，编译器类型将a和b强制转换为short，然后调用__mulhi3()。

我想知道如何更改gcc源代码[哪个文件]，这样我才能达到上述要求。

Answer 1

编译器可以分析代码并查看当您将结果立即转换为short时，多重复制可以作为short乘法进行，而不会影响结果。这与您的示例中的情况二完全相同。

结果相同，您不必担心使用哪个乘法函数。

在GNU GCC 4.5.3中输入int to short

1 个答案: