如何初始化char**？

Question

這是我的一段代碼：

char** filename;
*(filename) = "initialize";
printf("filename = %s",*(filename));

當我嘗試運行它時出現此錯誤：

Run-Time Check Failure #3 - The variable 'filename' is being used without being initialized.

有沒有什么辦法解決這一問題？

Answer 1

char *a  =  "abcdefg";
char **fileName = &a;

Answer 2

C方式：

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

char * filename = (char*) malloc( 100 ); // reserve 100 bytes in memory
strcpy(filename,"initialize");           // copy "initialize" into the memory
printf("filename = %s",filename);        // print out
free(filename);                          // free memory
filename = 0;                            // invalid pointers value is NULL

C++方式：

#include <string>
#include <iostream>

string filename("initialize");           // create string object
cout << "filename = " << filename;       // write out to stanard out

Answer 3

您需要使用 new 或 malloc 為文件名分配空間。 事實上，文件名只是一個指向 memory 你沒有請求的隨機區域的指針......

  filename = new char*;

Answer 4

char** filename = new char*;   
*(filename) = "initialize";    
printf("filename = %s",*(filename));

但是你為什么需要那些東西？

Answer 5

@Naszta 的答案是您應該聽的。 但是要糾正所有其他關於new的錯誤答案：

size_t len = strlen("initialize") + 1;
char* sz = new char [len];
strncpy(sz, "initialize", strlen("initialize"));

---

當然，真正的C++ 方法更好。

string filename = "initialize";
cout << "filename = " << filename;

Answer 6

您尚未分配要分配的 char* ：

char** filename = new char*;
*filename = "initialize";