c - 当 int 和 long 在 C 中具有相同的宽度时，排名有时会失败

Question

让平台 1 的宽度为 int 4 个字节和 long 的宽度8 个字节。

让平台 2 的宽度为 int 4 个字节和 long 的宽度与int的宽度相同.

然后给出:

unsigned int x = 2;
long signed int y = 3;
func(x * y);

在平台1上运行时，func的第一个参数的有效类型是long signed int .这符合预期。在 § 6.3.1.1.1.4 之后，unsigned int类型与 signed int 具有相同等级.然后它也遵循signed int type 的排名低于 long signed int根据 § 6.3.1.1.1.3。然后，这会触发乘法结果转换为 long signed int。类型，遵循 § 6.3.1.8.1.4.4。太棒了!

在平台2上运行时，乘法结果为long unsigned int . 为什么？

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背景

C99 标准第 6.3.1.1 节第 1 小节第 3 点说:

The rank for long long int shall be greater than the rank of long int,which shall be greater than the rank of int, which shall be greaterthat the rank of short int, which shall be greater than the rank ofsigned char.

这表明 long int排名高于 int .

另外，C99标准同段第4点说:

The rank of any unsigned integer type shall equal the rank of thecorresponding signed integer type, if any.

这里有几件事是 unsigned int类型与 signed int 具有相同等级类型。同样，long unsigned int类型与 long signed int 具有相同等级.

最后，在 § 6.3.1.8 节的第 1 小节中，第 4.3 点说:

Otherwise, if the operand that has unsigned integer type has rankgreater or equal to the rank of the type of the other operand, thenthe operand with signed integer type is converted to the type of theoperand with unsigned integer type.

和第 4.4 点:

Otherwise, if the type of the operand with signed integer type canrepresent all values of the type of the operand with unsigned integertype, then the operand with unsigned integer type is converted to thetype of the operand with signed integer type.

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测试代码

#include <stdio.h>

#define func(x) _Generic((x), long unsigned int: func_longunsignedint, long signed int: func_longsignedint, signed int: func_signedint, unsigned int: func_unsignedint)(x);

void func_longunsignedint (long unsigned int x)
{
    printf("%s\t%lu\n", __func__, x);
}

void func_longsignedint (long signed int x)
{
    printf("%s\t%ld\n", __func__, x);
}

void func_signedint (signed int x)
{
    printf("%s\t%d\n", __func__, x);
}

void func_unsignedint (unsigned int x)
{
    printf("%s\t%u\n", __func__, x);
}

int main(void)
{
    printf("int width %d\n", sizeof(int));
    printf("long width %d\n", sizeof(long));
    unsigned int x = 2;
    long signed int y = -3;
    func(x * y);
}

对于平台 1，使用 gcc -m64 编译希望将 long 强制为 8 个字节，将 int 强制为 4 个字节。

对于平台 2，使用 gcc -m32 编译希望将 long 强制为 4 个字节，并将 int 强制为 4 个字节。

Answer 1

我们采用unsigned 和long signed int 并遵循C11 6.3.1.8 :

• Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsignedinteger type, then the operand with unsigned integer type is convertedto the type of the operand with signed integer type.

在 64 位平台 #1 类型 long signed int 可以表示 unsigned 的所有值，因为它有 64 位 vs unsigned 有32位。所以 unsigned 被转换为 long signed int。

在 32 位平台 #2 上，类型 long signed int 不能表示 unsigned 的所有值(UINT_MAX=2^32 但LONG_MAX=2^31-1)。所以我们继续...

• Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

...所以在平台 #2 上，“无符号整数类型对应于有符号整数类型”——这是 long signed int + unsigned = long unsigned int。因此在平台 #2 上，两个操作数都被转换为 long unsigned int，因此是您看到的结果。