为什么下列程序进行的是无符号乘法？

This topic created in 927 days ago, the information mentioned may be changed or developed.

unsigned int x1 = 0xaeb1c2aa;
unsigned int x2 = 0xaeb1c2aa;

long long r_whole = (signed long long)x1 * (signed long long)x2;

printf("r_whole = %lld\n",r_whole);

输出的结果是无符号乘法的结果：

r_whole = 8590088583138384100

而下面的程序

unsigned int x1 = 0xaeb1c2aa;
unsigned int x2 = 0xaeb1c2aa;

int xx1 = x1;
int xx2 = x2;

long long r_whole = (signed long long)xx1 * (signed long long)xx2;

printf("r_whole = %lld\n",r_whole);

输出的事有符号乘法的结果：

r_whole = 1860719719092984036

第一段程序里我不是对 x1 进行强制类型转换了吗

unsigned

signed

conversion

6 replies • 2024-04-20 14:32:11 +08:00

chingyat

Apr 19, 2024 via iPhone

因为后者 xx1 和 xx2 是负数.

Hsinyao

Apr 19, 2024

不要纠结这些语法，建议 objdump 看汇编

lindt99cocoa

Apr 19, 2024

无符号数做零扩展，有符号数做符号扩展

ysc3839

Apr 19, 2024 via Android

因为 1 没符号，转成有符号类型时也不带符号，2 有符号，转成有符号类型时就会带符号

cnbatch

Apr 19, 2024

写成这样就容易懂了：

uint32_t x1 = 0xaeb1c2aa; // 2930885290
uint32_t x2 = 0xaeb1c2aa;

int64_t var1 = x1;
int64_t var2 = x2;

long long r_whole = var1 * var2;

x1 和 x2 强制转换成 signed long long 的时候，实际上就相当于 int64_t var1 = x1;
有符号 long long 能够容纳的范围足够大，大到可以完整“吞下”整个 0xaeb1c2aa

iceheart

Apr 20, 2024 via Android

第二个数值溢出，转成 int 型变成负数了
int 扩展成 long long ，高位按符号位扩展，所以仍是负数。