C语言 printf 是否需要 %zu 说明符？
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Is the %zu specifier required for printf?
提问by Trevor Hickey
We are using C89 on an embedded platform. I attempted to print out a size_t, but it did not work:
我们在嵌入式平台上使用 C89。我试图打印出一个size_t，但没有奏效：
#include <stdio.h>
int main(void) {
size_t n = 123;
printf("%zu\n",n);
return 0;
}
Instead of 123, I got zu.
Other specifiers work correctly.
而不是123，我得到了zu。
其他说明符正常工作。
If size_texists shouldn't zualso be available in printf?
Is this something I should contact my library vendor about, or is a library implementation allowed to exclude it?
如果size_t存在也不应该 zu在printf?
这是我应该联系我的库供应商的事情，还是允许库实现排除它？
回答by P.P
If size_t exists shouldn't zu also be available in printf?
如果 size_t 存在，则 zu 不应该在 printf 中可用吗？
size_texisted at least since C89 but the respective format specifier %zu(specifically the length modifier z) was added to the standard only since C99.
size_t至少自 C89 以来就存在，但仅自 C99 以来才将相应的格式说明符%zu（特别是长度修饰符z）添加到标准中。
So, if you can't use C99 (or C11) and had to print size_tin C89, you just have to fallback to other existing types, such as:
因此，如果您不能使用 C99（或 C11）并且不得不size_t在 C89 中打印，您只需要回退到其他现有类型，例如：
printf("%lu\n", (unsigned long)n);
回答by Praveen Kumar Rana
This is working fine. In my system, it's printing the expected value. You may be getting the error due to the compiler. I am using the GCCcompiler. If you have this only and trying to get the output the first update the GCC and then try. It will work.
这工作正常。在我的系统中，它正在打印预期值。由于编译器，您可能会收到错误。我正在使用GCC编译器。如果您只有这个并尝试获取输出，请首先更新 GCC，然后尝试。它会起作用。
