c - 如何使用 _spawn 或 _exec 进行引导？

Question

编写以下程序后，它似乎没有将参数传递给被调用的应用程序。研究时_spawnv以及它能做什么，_execvp被认为是一个合适的替代方案。有谁在源代码中看到问题并知道需要做什么来解决它？

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

int main(int argc, char** argv)
{
    int index;
    char** args;
    args = (char**) malloc((argc + 1) * sizeof(char*));
    args[0] = "boids.py";
    for (index = 1; index < argc; index++)
    {
        args[index - 1] = argv[index];
    }
    args[argc] = NULL;
    return _execvp("python", args);
}

Answer 1

argv vector 中的第一个参数通常是要启动的可执行文件的完全限定名称:

The _spawnv, _spawnve, _spawnvp, and _spawnvpe calls are useful when there is a variable number of arguments to the new process. Pointers to the arguments are passed as an array, argv. The argument argv[0] is usually a pointer to a path in real mode or to the program name in protected mode, and argv1 through argv[n] are pointers to the character strings forming the new argument list. The argument argv[n +1] must be a NULL pointer to mark the end of the argument list.

(来自 MSDN )

同样:

The _execv, _execve, _execvp, and _execvpe calls are useful when the number of parameters to the new process is variable. Pointers to the parameters are passed as an array, argv. The parameter argv[0] is usually a pointer to cmdname. The parameters argv1 through argv[n] point to the character strings forming the new parameter list. The parameter argv[n+1] must be a NULL pointer to mark the end of the parameter list.

( MSDN )