c++ - 无法使用 C++ stdlib 系统调用运行 make

Question

我在 C++ 中得到了以下代码

  if (should_run_make) {
    std::string make = "make -C ";
    make.append(outdir);
    std::cout << "Make cmd is " << make << std::endl;
    system(make.c_str());
  }

报告如下:

Make cmd is make -C /home/hamiltont/temp/ make: Entering directory /home/hamiltont/temp' make: *** No targets. Stop.
make: Leaving directory/home/hamiltont/temp'

但是，以多种方式手动执行此操作效果很好，例如

[hamiltont@4 generator]$ make -C /home/hamiltont/temp/
make: Entering directory `/home/hamiltont/temp'
g++ -O3 -I/usr/include/openmpi-x86_64 -L/usr/local/lib -L/usr/lib64/openmpi/lib -lmpi -lmpi_cxx -lboost_serialization -lboost_mpi  stg_impl.cpp -o impl
make: Leaving directory `/home/hamiltont/temp'

[hamiltont@4 generator]$ cd /home/hamiltont/temp/
[hamiltont@4 temp]$ make
g++ -O3 -I/usr/include/openmpi-x86_64 -L/usr/local/lib -L/usr/lib64/openmpi/lib -lmpi -lmpi_cxx -lboost_serialization -lboost_mpi  stg_impl.cpp -o impl

Answer 1

您是从 C 程序中生成联编文件吗？这是我能想到的唯一会导致特定错误消息的原因。

make: *** No targets. Stop. 

Reproducing the error

Here's how I could generate that message:

#include <stdio.h>
#include <stdlib.h>
int main()
{
    FILE *fp = fopen("Makefile", "w");
    fputs("all:\n\techo Done.\n", fp);
    system("make");
    fclose(fp);
    return 0;
}

这可以预见地打印:

make: *** No targets.  Stop.

I say predictably because Makefile will be empty! This is because IO is buffered...

Fixed version

So, I close the file before calling system(), which flushes the buffer (fflush() would also do the trick):

#include <stdio.h>
#include <stdlib.h>
int main()
{
    FILE *fp = fopen("Makefile", "w");
    fputs("all:\n\techo Done.\n", fp);
    fclose(fp);
    system("make");
    return 0;
}

输出:

echo Done.Done.

为了清楚起见，我使用了 C 的 IO 函数，但同样的规则适用于 <iostream> .