如何在c中打印指针变量

问题描述

现在，我知道这看起来像是早期查询的重复，但我已经搜索并尝试了所有内容。我无法打印双指针。我有以下代码

Now, I understand this would look like a repeat of earlier queries, but I have searched and tried everything. I am not able to print a double pointer. I have the following code

int main(void)
{

char *line;
  char **args;
  int status;

  do {
    //stuff
    args = split(&line);

    printf(*args);
}

现在一切正常，只是在最后的打印行中，我收到以下错误：

Now it is all executing fine, Just that in the final print line, I am getting the below error:

expected 'const char * __restrict__' but argument is of type 'char ***'
 extern int printf (const char *__restrict __format, ...);

现在，我知道为什么我收到这个错误，因为我试图非法打印一些东西。我只是不知道我需要修复到底是什么，如果我需要打印这些双指针，我们该怎么做呢。



谢谢



我尝试过：



我研究过类似的答案，但没有真正解决我的问题，我也尝试阅读关于指针的文献，但函数split（）的返回值是一个指针，我不能改变它。

Now, I know why I am getting this error, because I am trying to print something illegally. I just don't know what exactly is it that I need to fix, also, if I ever need to print these double pointers, how do we go about it.

Thanks

What I have tried:

I have researched for similar answers, but nothing really solves my problem, I have also tried to read literature on pointers, but the return value of the function split() is a pointer and I cannot change that.

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