C - 将 'string' 而不是 char 分配给 open 函数使用的变量

Question

我有一个由以下内容组成的变量输出:

//main function
char *output; //this is how output is defined

//these 3 lines are in some function call
char cmd[50]  // this is how cmd is defined
cmd = "test.txt" //cmd is not initialized this way (cmd = "text.txt"). 
                 //Rather, there is some function call that passes in the 
                 //&output and does some things, inevitably updating cmd 
                 //by adding one character at a time. In this example it will 
                 //go cmd = "t", cmd = "te", ... cmd = "test.txt",
                 //cmd = "test.txt"
*output = cmd; //then at the end of the function I try to set the contents of 
               //output to the cmd.

然后我尝试调用 open 方法:

int outfile;
outfile = open( *output, O_WRONLY | O_CREAT, S_IRUSR | S_IWUSR | S_IRGRP );

但这不起作用。然而这有效:

int outfile;
outfile = open( "test.txt", O_WRONLY | O_CREAT, S_IRUSR | S_IWUSR | S_IRGRP );

当我尝试传递 *output 时，出现错误:

warning: incompatible integer to pointer conversion passing 'char' to parameter of type 'const char *'; remove *

但是当我尝试仅输出时，它无法正常工作。我假设输出中存储的内容只是“t”，但我需要所有这些才能使这项工作正常进行。有任何想法吗？其背景是我正在手动尝试进行重定向。

Answer 1

*输出是 - 你猜对了 - 一个字符(因为输出是“ptr to char”)cmd 是 - 你没有猜到 - 一个指向 char 的指针(因为它是一个数组)

所以你的赋值“*output = cmd”实际上不会做任何合理的事情。

尝试

strcpy(output, cmd);

或者，(更好)

strncpy(output, cmd, BUFLEN);

(确保 BUFLEN 等于输出时分配的内存，并确保输出不是局部变量)

托比亚斯