在mingw-w64 gcc 7.1中，如何在没有警告的情况下打印尺寸？

Question

我正在使用minGW的minw-w64（x64）分叉nuwen.net.这是来自gcc的7.1版本：

gcc --version
gcc (GCC) 7.1.0

我正在编译这个程序：

#include <stdio.h>

int main(void)
{
    size_t a = 100;
    printf("a=%lu\n",a);
    printf("a=%llu\n",a);
    printf("a=%zu\n",a);
    printf("a=%I64u\n",a);
}

带有警告和c11标准：

gcc -Wall -Wextra -Wpedantic -std=c11 test_size_t.c

我得到了这些警告：

   test_size_t.c: In function 'main':
    test_size_t.c:6:14: warning: format '%lu' expects argument of type 'long unsigned int', but argument 2 has type 'size_t {aka long long unsigned int}' [-Wformat=]
      printf("a=%lu\n",a);
                ~~^
                %I64u
    test_size_t.c:6:14: warning: format '%lu' expects argument of type 'long unsigned int', but argument 2 has type 'size_t {aka long long unsigned int}' [-Wformat=]
      printf("a=%lu\n",a);
                ~~^
                %I64u
    test_size_t.c:7:14: warning: unknown conversion type character 'l' in format [-Wformat=]
      printf("a=%llu\n",a);
                  ^
    test_size_t.c:7:9: warning: too many arguments for format [-Wformat-extra-args]
      printf("a=%llu\n",a);
             ^~~~~~~~~~
    test_size_t.c:7:14: warning: unknown conversion type character 'l' in format [-Wformat=]
      printf("a=%llu\n",a);
                  ^
    test_size_t.c:7:9: warning: too many arguments for format [-Wformat-extra-args]
      printf("a=%llu\n",a);
             ^~~~~~~~~~
    test_size_t.c:8:13: warning: unknown conversion type character 'z' in format [-Wformat=]
      printf("a=%zu\n",a);
                 ^
    test_size_t.c:8:9: warning: too many arguments for format [-Wformat-extra-args]
      printf("a=%zu\n",a);
             ^~~~~~~~~
    test_size_t.c:8:13: warning: unknown conversion type character 'z' in format [-Wformat=]
      printf("a=%zu\n",a);
                 ^
    test_size_t.c:8:9: warning: too many arguments for format [-Wformat-extra-args]
      printf("a=%zu\n",a);
             ^~~~~~~~~
    test_size_t.c:9:9: warning: ISO C does not support the 'I64' ms_printf length modifier [-Wformat=]
      printf("a=%I64u\n",a);
         ^~~~~~~~~~~
    test_size_t.c:9:9: warning: ISO C does not support the 'I64' ms_printf length modifier [-Wformat=]

我想在没有警告的情况下打印size_t，但在这种情况下不知道正确的格式说明符。

Answer 1

PRIuPTRtrick-works（类型size\u t？）变量的跨平台格式字符串：

#include <stdio.h>
#include <inttypes.h>
int main()
{
    size_t a = (size_t)(1024ULL * 1024 * 1024 * 1024 + 13);
    printf("a                 = %" PRIuPTR "\n", a);
    printf("sizeof(size_t)    = %" PRIuPTR "\n", sizeof(size_t));
    printf("sizeof(uintptr_t) = %" PRIuPTR "\n", sizeof(uintptr_t));
    return 0;
}

输出x86：

a                 = 13
sizeof(size_t)    = 4
sizeof(uintptr_t) = 4

输出x64：

a                 = 1099511627789
sizeof(size_t)    = 8
sizeof(uintptr_t) = 8

Answer 2

注释中提到的替代解决方案是加入\u USE\u MINGW\u ANSI\u STDIO编译器开关：

#define __USE_MINGW_ANSI_STDIO 1

#include <stdio.h>

int main(void)
{
    size_t a = 100;
    printf("a=%lu\n",a);
    printf("a=%llu\n",a);
    printf("a=%zu\n",a);
    printf("a=%I64u\n",a);
}

这会使代码按预期编译，gcc现在会给出相应的警告：

warning: format '%lu' expects argument of type 'long unsigned int', but argument 2 has type 'size_t' [-Wformat=]  
warning: ISO C does not support the 'I' printf flag [-Wformat=]  
warning: format '%u' expects argument of type 'unsigned int', but argument 2 has type 'size_t' [-Wformat=]  

或者，您可以在命令行中使用-D__USE_MINGW_ANSI_STDIO=1定义宏。

Answer 3

问题不在于编译器，而在于C库。MinGW使用Microsoft的“Visual C运行时”（msvcrt），它只符合c89，不支持z格式说明符。

以下是使用MinGW时安全打印大小\u t的方法：

#include <inttypes.h>
#include <stdio.h>

#ifdef _WIN32
#  ifdef _WIN64
#    define PRI_SIZET PRIu64
#  else
#    define PRI_SIZET PRIu32
#  endif
#else
#  define PRI_SIZET "zu"
#endif

int main(void)
{
    size_t mySize = 24;

    printf("%" PRI_SIZET "\n", mySize);
}

在win64上，您会收到此代码的警告，因为PRIu64扩展到msvcrt特定的I64u格式说明符。但是您可以使用GCC标志-Wno-pedian-ms-form来静音此警告。

请注意，对于long（此处在32位和64位窗口上使用PRIu64），您需要一个类似的技巧，因为msvcrt也不知道ll。

编辑：正如@M.M在评论中指出的，您可以将MinGW提供的支持C11的可选stdio函数链接到#define uu使用MinGW\uANSI_Stdio1。如果我能绕过msvcrt的特性，我宁愿不链接额外的代码，但这当然是一个品味问题。