如何在 C 中便携式打印 int64_t 类型

发布于 2025-01-03 22:16:29 字数 1099 浏览 2 评论 0原文

C99 标准具有字节大小的整数类型，如 int64_t。我当前使用 Windows 的 %I64d 格式（或无符号 %I64u），例如：

#include <stdio.h>
#include <stdint.h>
int64_t my_int = 999999999999999999;
printf("This is my_int: %I64d\n", my_int);

我收到此编译器警告：

warning: format ‘%I64d’ expects type ‘int’, but argument 2 has type ‘int64_t’

我尝试过：

printf("This is my_int: %lld\n", my_int); // long long decimal

但我收到相同的警告。我正在使用这个编译器：

~/dev/c$ cc -v
Using built-in specs.
Target: i686-apple-darwin10
Configured with: /var/tmp/gcc/gcc-5664~89/src/configure --disable-checking --enable-werror --prefix=/usr --mandir=/share/man --enable-languages=c,objc,c++,obj-c++ --program-transform-name=/^[cg][^.-]*$/s/$/-4.2/ --with-slibdir=/usr/lib --build=i686-apple-darwin10 --program-prefix=i686-apple-darwin10- --host=x86_64-apple-darwin10 --target=i686-apple-darwin10 --with-gxx-include-dir=/include/c++/4.2.1
Thread model: posix
gcc version 4.2.1 (Apple Inc. build 5664)

我应该使用哪种格式来打印 my_int 变量而不发出警告？

原文

C99 standard has integer types with bytes size like int64_t. I am using Windows's %I64d format currently (or unsigned %I64u), like:

#include <stdio.h>
#include <stdint.h>
int64_t my_int = 999999999999999999;
printf("This is my_int: %I64d\n", my_int);

and I get this compiler warning:

warning: format ‘%I64d’ expects type ‘int’, but argument 2 has type ‘int64_t’

I tried with:

printf("This is my_int: %lld\n", my_int); // long long decimal

But I get the same warning. I am using this compiler:

~/dev/c$ cc -v
Using built-in specs.
Target: i686-apple-darwin10
Configured with: /var/tmp/gcc/gcc-5664~89/src/configure --disable-checking --enable-werror --prefix=/usr --mandir=/share/man --enable-languages=c,objc,c++,obj-c++ --program-transform-name=/^[cg][^.-]*$/s/$/-4.2/ --with-slibdir=/usr/lib --build=i686-apple-darwin10 --program-prefix=i686-apple-darwin10- --host=x86_64-apple-darwin10 --target=i686-apple-darwin10 --with-gxx-include-dir=/include/c++/4.2.1
Thread model: posix
gcc version 4.2.1 (Apple Inc. build 5664)

Which format should I use to print my_int variable without having a warning?

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箜明 2025-01-10 22:16:29

对于 int64_t 类型：

#include <inttypes.h>
int64_t t;
printf("%" PRId64 "\n", t);

对于 uint64_t 类型：

#include <inttypes.h>
uint64_t t;
printf("%" PRIu64 "\n", t);

您还可以使用 PRIx64 以十六进制打印。

cppreference.com 提供了所有类型的可用宏的完整列表，包括 <代码>intptr_t（PRIxPTR）。 scanf 有单独的宏，例如 SCNd64。

PRIu16 的典型定义是 "hu"，因此隐式字符串常量连接发生在编译时。

为了使您的代码完全可移植，您必须使用 PRId32 等来打印 int32_t，并使用 "%d" 或类似的方式来打印 <代码>int。

For int64_t type:

#include <inttypes.h>
int64_t t;
printf("%" PRId64 "\n", t);

for uint64_t type:

#include <inttypes.h>
uint64_t t;
printf("%" PRIu64 "\n", t);

you can also use PRIx64 to print in hexadecimal.

cppreference.com has a full listing of available macros for all types including intptr_t (PRIxPTR). There are separate macros for scanf, like SCNd64.

A typical definition of PRIu16 would be "hu", so implicit string-constant concatenation happens at compile time.

For your code to be fully portable, you must use PRId32 and so on for printing int32_t, and "%d" or similar for printing int.

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暗藏城府 2025-01-10 22:16:29

C99 的方式是

#include <inttypes.h>
int64_t my_int = 999999999999999999;
printf("%" PRId64 "\n", my_int);

或者你可以铸造！

printf("%ld", (long)my_int);
printf("%lld", (long long)my_int); /* C89 didn't define `long long` */
printf("%f", (double)my_int);

如果您坚持使用 C89 实现（尤其是 Visual Studio），您也许可以使用开源（和）：http://code.google.com/p/msinttypes/

The C99 way is

#include <inttypes.h>
int64_t my_int = 999999999999999999;
printf("%" PRId64 "\n", my_int);

Or you could cast!

printf("%ld", (long)my_int);
printf("%lld", (long long)my_int); /* C89 didn't define `long long` */
printf("%f", (double)my_int);

If you're stuck with a C89 implementation (notably Visual Studio) you can perhaps use an open source <inttypes.h> (and <stdint.h>): http://code.google.com/p/msinttypes/

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北城挽邺 2025-01-10 22:16:29

在 C99 中，%j 长度修饰符还可以与 printf 系列函数一起使用，以打印 int64_t 和 uint64_t 类型的值：

#include <stdio.h>
#include <stdint.h>

int main(int argc, char *argv[])
{
    int64_t  a = 1LL << 63;
    uint64_t b = 1ULL << 63;

    printf("a=%jd (0x%jx)\n", a, a);
    printf("b=%ju (0x%jx)\n", b, b);

    return 0;
}

编译此代码使用 gcc -Wall -pedantic -std=c99 不会产生警告，并且程序会打印预期的输出：

a=-9223372036854775808 (0x8000000000000000)
b=9223372036854775808 (0x8000000000000000)

这是根据我的 Linux 系统上的 printf(3) （手册页明确指出，j 用于指示到 intmax_t 或 uintmax_t 的转换；在我的 stdint.h 中，两者都 int64_t 和 intmax_t 的类型定义方式完全相同，uint64_t 也类似）。我不确定这是否可以完美移植到其他系统。

With C99 the %j length modifier can also be used with the printf family of functions to print values of type int64_t and uint64_t:

#include <stdio.h>
#include <stdint.h>

int main(int argc, char *argv[])
{
    int64_t  a = 1LL << 63;
    uint64_t b = 1ULL << 63;

    printf("a=%jd (0x%jx)\n", a, a);
    printf("b=%ju (0x%jx)\n", b, b);

    return 0;
}

Compiling this code with gcc -Wall -pedantic -std=c99 produces no warnings, and the program prints the expected output:

a=-9223372036854775808 (0x8000000000000000)
b=9223372036854775808 (0x8000000000000000)

This is according to printf(3) on my Linux system (the man page specifically says that j is used to indicate a conversion to an intmax_t or uintmax_t; in my stdint.h, both int64_t and intmax_t are typedef'd in exactly the same way, and similarly for uint64_t). I'm not sure if this is perfectly portable to other systems.

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琉璃繁缕 2025-01-10 22:16:29

来自嵌入式世界，甚至 uclibc 也并不总是可用，并且代码类似于

uint64_t myval = 0xdeadfacedeadbeef;
printf("%llx", myval);

正在打印你的垃圾或者根本不起作用——我总是使用一个小助手，它允许我正确转储 uint64_t 十六进制：

#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>

char* ullx(uint64_t val)
{
    static char buf[34] = { [0 ... 33] = 0 };
    char* out = &buf[33];
    uint64_t hval = val;
    unsigned int hbase = 16;

    do {
        *out = "0123456789abcdef"[hval % hbase];
        --out;
        hval /= hbase;
    } while(hval);

    *out-- = 'x', *out = '0';

    return out;
}

Coming from the embedded world, where even uclibc is not always available, and code like

uint64_t myval = 0xdeadfacedeadbeef; printf("%llx", myval);

is printing you crap or not working at all -- i always use a tiny helper, that allows me to dump properly uint64_t hex:

#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>

char* ullx(uint64_t val)
{
    static char buf[34] = { [0 ... 33] = 0 };
    char* out = &buf[33];
    uint64_t hval = val;
    unsigned int hbase = 16;

    do {
        *out = "0123456789abcdef"[hval % hbase];
        --out;
        hval /= hbase;
    } while(hval);

    *out-- = 'x', *out = '0';

    return out;
}

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情仇皆在手 2025-01-10 22:16:29

在windows环境下使用，

%I64d

在Linux下使用

%lld

In windows environment, use

%I64d

in Linux, use

%lld

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丧 2025-01-10 22:16:29

当我寻找一种以十六进制显示 64 位数字的方法时，偶然发现了这个问题：

发现您可以使用：

0x%016llx - 至少适用于我的编译器（aarch64 GCC 7.3 .0)

回复收藏 0 原文

夏夜暖风 2025-01-10 22:16:29

//VC6.0（386 及更好）

    __int64 my_qw_var = 0x1234567890abcdef;

    __int32 v_dw_h;
    __int32 v_dw_l;

    __asm
        {
            mov eax,[dword ptr my_qw_var + 4]   //dwh
            mov [dword ptr v_dw_h],eax

            mov eax,[dword ptr my_qw_var]   //dwl
            mov [dword ptr v_dw_l],eax

        }
        //Oops 0.8 format
    printf("val = 0x%0.8x%0.8x\n", (__int32)v_dw_h, (__int32)v_dw_l);

问候。

//VC6.0 (386 & better)

    __int64 my_qw_var = 0x1234567890abcdef;

    __int32 v_dw_h;
    __int32 v_dw_l;

    __asm
        {
            mov eax,[dword ptr my_qw_var + 4]   //dwh
            mov [dword ptr v_dw_h],eax

            mov eax,[dword ptr my_qw_var]   //dwl
            mov [dword ptr v_dw_l],eax

        }
        //Oops 0.8 format
    printf("val = 0x%0.8x%0.8x\n", (__int32)v_dw_h, (__int32)v_dw_l);

Regards.

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