gcc - 在 GNU GCC 4.5.3 中，将 int 类型转换为 short

Question

我想知道，当两个整数相乘并将结果类型转换为short并分配给short时，编译器会将其解析为什么？下面是代码片段

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 for 的情况下q =(short)(a*b);，应该执行前两个整数乘法（使用__mulsi3()），然后将其分配给 short。但这里不是这种情况，编译器类型将两者都转换为a和b，然后调用__mulhi3().

我想知道如何更改 gcc 源代码 [哪个文件]，以便我可以实现上述要求。

Answer 1

编译器可以分析代码并看到当您立即将结果转换为 ashort时，可以将乘法作为乘法来完成，short而不会影响结果。这与您的示例的情况二完全相同。

结果是一样的，你不必担心使用哪个乘法函数。