c++11 - 如何在 g++ -std=c++11 version 4.8.2 中使用 snprintf()

Question

我正在尝试使用 -std=c++11 编译在 4.8.2 版 g++ 中使用 snprintf 的代码，但没有成功。
为什么 g++ 不识别 snprintf？我可以在仍然使用 snprintf 和 c++11 的同时克服这个问题吗？

我得到以下信息:

make all  Building file: ../src/cppHWtest.cpp Invoking: Cygwin C++
Compiler g++ -std=c++11 -D"hash_map=unordered_map" -O0 -g3 -Wall -c
-fmessage-length=0 -MMD -MP -MF"src/cppHWtest.d" -MT"src/cppHWtest.d" -o "src/cppHWtest.o" "../src/cppHWtest.cpp" cygwin warning:   MS-DOS style path detected: C:\Users\poudyal\workspace\cppHWtest\Debug
Preferred POSIX equivalent is:
/cygdrive/c/Users/poudyal/workspace/cppHWtest/Debug   CYGWIN
environment variable option "nodosfilewarning" turns off this warning.
Consult the user's guide for more details about POSIX paths:
    http://cygwin.com/cygwin-ug-net/using.html#using-pathnames ../src/cppHWtest.cpp: In function 'int main()':
../src/cppHWtest.cpp:20:35: error: 'snprintf' was not declared in this
scope   snprintf(buff, 10, "%s", "Hell O");
                                   ^ make: *** [src/cppHWtest.o] Error 1 src/subdir.mk:18: recipe for target 'src/cppHWtest.o' failed

**** Build Finished ****

Answer 1

可能比将模式切换到 -std=gnu++11 侵入性更小的解决方案是 -U__STRICT_ANSI__ .这将启用 #if !defined(__STRICT_ANSI__) || (__STDC_VERSION__ >= 199901L)在 stdio.h:234 (GCC 4.8.3)。