问题描述
我有一个 size_t 类型的变量，我想使用 printf() 打印它.我使用什么格式说明符来便携地打印它?
I have a variable of type size_t, and I want to print it using printf(). What format specifier do I use to print it portably?
在 32 位机器中，%u 似乎是正确的.我用 g++ -g -W -Wall -Werror -ansi -pedantic 编译，没有警告.但是当我在 64 位机器上编译该代码时，它会产生警告.
In 32-bit machine, %u seems right. I compiled with g++ -g -W -Wall -Werror -ansi -pedantic, and there was no warning. But when I compile that code in 64-bit machine, it produces warning.
size_t x = <something>;
printf("size = %u
", x);
warning: format '%u' expects type 'unsigned int',
but argument 2 has type 'long unsigned int'
正如预期的那样，如果我将其更改为 %lu，警告就会消失.
The warning goes away, as expected, if I change that to %lu.
问题是，我该如何编写代码，才能在 32 位和 64 位机器上编译时无警告?
The question is, how can I write the code, so that it compiles warning free on both 32- and 64- bit machines?
作为一种解决方法，我想一个答案可能是将变量转换"为一个足够大的整数，例如 unsigned long，然后使用 %lu.这在两种情况下都有效.我正在寻找是否有任何其他想法.
As a workaround, I guess one answer might be to "cast" the variable into an integer that is big enough, say unsigned long, and print using %lu. That would work in both cases. I am looking if there is any other idea.
推荐答案
使用 z 修饰符:
size_t x = ...;
ssize_t y = ...;
printf("%zu
", x); // prints as unsigned decimal
printf("%zx
", x); // prints as hex
printf("%zd
", y); // prints as signed decimal
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