我不小心让argc和argv在main函数之外工作.
以下是重现此"错误"的代码(与GCC一起编译):
手C
#include <stdio.h>
int test(void);
int main(void)
{
test();
}
Test.c
#include <stdio.h>
int test(int argc, char **argv)
{
printf("argc = %i\n", argc);
printf("args:\n");
for (int i = 0; i < argc; i++)
{
printf("- %s\n", argv[i]);
}
printf("test called\n");
return 0;
}
As comments already hinted, it is not so much a case of
"I accidentally made argc and argv work",
but more of
"I made argc and argv work accidentally".
I.e. you did not stumble across something which works reliable,
instead your code only accidentally works in no reliable way.
The concept of "undefined behaviour" is an important one in C.
It pays to make yourself familiar with it. That can even be fun, e.g. read up on "nasal demons".
Most relevant for your case is that UB can, among other things, fiendishly deceive you into believing that your program works; I consider this to actually be the worst case of possible outcomes.
正如pmacfarlane在 comments 中给出的原因，你最终会有不确定的行为:
您得到了未定义的行为，因为您使用错误的参数数量和类型调用了test().