在比较UINT16_T和UNSIGNED INT时，为什么存在签名问题？
我有这样的代码：
#include <iostream>
using std::cout;
using std::endl;
int main() {
uint16_t a = 0;
uint16_t b = 0;
if ( a - b < 3u )
{
cout << "cacahuète" << endl;
}
return 0;
}
当我使用-wall对G ++进行编译时，我会得到：
temp.cpp: In function ‘int main()’:
temp.cpp:9:13: warning: comparison of integer expressions of different signedness: ‘int’ and ‘unsigned int’ [-Wsign-compare]
9 | if ( a - b < 3u )
| ~~~~~~^~~~
如果我编写，则不会显示该警告，如果（a -b＆lt; static_cast＆static_cast＆lt; uint16_t＆gt ;（3U））而不是。
- 那是怎么回事？ INT来自哪里？
- 这实际上会导致不正确的行为吗？
- 是否有较少的详细方法来使它保持沉默？ （还是写uint16_t字面的详细方法？）
I have a code like this :
#include <iostream>
using std::cout;
using std::endl;
int main() {
uint16_t a = 0;
uint16_t b = 0;
if ( a - b < 3u )
{
cout << "cacahuète" << endl;
}
return 0;
}
When I compile it using g++ with -Wall, I get :
temp.cpp: In function ‘int main()’:
temp.cpp:9:13: warning: comparison of integer expressions of different signedness: ‘int’ and ‘unsigned int’ [-Wsign-compare]
9 | if ( a - b < 3u )
| ~~~~~~^~~~
That warning doesn't show up if I write if ( a - b < static_cast<uint16_t>(3u) ) instead.
- So what's going on, here ? Where does the int come from?
- Can this actually result in an incorrect behavior?
- Is there a less verbose way to silence it? (or a less verbose way to write a uint16_t literal?)
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整数促销正在这里进行。在
std :: uint16_t的系统上，它小于int，将其升级为int（大多数二进制操作）时。在
a -b中，两个操作数均升级为int，结果是int。您将此签名的整数与3U进行比较，该整数是insted int。迹象有所不同，因为编译器警告您。在这里，右手操作数也被提升为
int。比较的两面都签署了，因此没有警告。如果（a -b＆lt; static_cast＆lt; uint16_t＆gt;（3U））的行为确实不同于a -b＆lt; static_cast＆lt; uint16_t＆gt;（3U）。如果一个是正确的，那么大概另一个是不正确的。正确的解决方案取决于您想要正确的行为。
ps您忘了包含定义
uint16_t的标题。Integer promotion is going on here. On systems where
std::uint16_tis smaller thanint, it will be promoted tointwhen used as an operand (of most binary operations).In
a - bboth operands are promoted tointand the result isintalso. You compare this signed integer to3uwhich isunsigned int. The signs differ, as the compiler warns you.Here, the right hand operand is also promoted to
int. Both sides of comparison are signed so there is no warning.if ( a - b < static_cast<uint16_t>(3u) )does have different behaviour thana - b < static_cast<uint16_t>(3u). If one is correct, then presumably the other is incorrect.The correct solution depends on what behaviour you want to be correct.
P.S. You forgot to include the header that defines
uint16_t.