问题描述
我收到以下代码的分段错误.有人可以解释为什么吗?我希望能够将 argv 的内容复制到一个名为 rArray 的新数组中.
I'm getting a segmentation fault for the following code. Can somebody explain why? I would like to be able to copy the contents of argv into a new array, which I called rArray.
#include <iostream>
using namespace std;
int main( int argc, char **argv)
{
char **rArray;
int numRows = argc;
cout << "You have " << argc << " arguments:" << endl << endl;
cout << "ARGV ARRAY" << endl;
for (int i = 0; i < argc; i++)
{
cout << argv[i] << endl;
}
cout << endl << endl << "COPIED ARRAY" << endl;
for(int i; i < numRows; i++)
{
for (int j = 0; j < argc; j++)
{
rArray[i][j] = argv[i][j];
}
}
for (int i = 0; i < argc; i++)
{
cout << "Copied array at index " << i << "is equal to " << rArray[i] << endl;;
}
cin.get();
}
程序输出:
/a.out hello world
You have 3 arguments:
ARGV ARRAY
./a.out
hello
world
COPIED ARRAY
Segmentation fault: 11
为什么我会收到这个错误?我该如何解决?
Why am I getting this error? How do I fix it?
我得到了一个修复,将 char **rArray
更改为 string rArray
,并从那里动态分配大小.
I got a fix, changing the char **rArray
to string rArray
, and dynamically allocating the size from there.
推荐答案
你需要为rArray
分配内存,还需要初始化外循环计数器i
.
You need to allocate memory for rArray
and also need to initialise the outer loop counter i
.
由于argv
的内容是常量字符串,你可以直接复制指向它们的指针
Since the contents of argv
are constant strings, you could just copy pointers to them
rArray = new char*[argc+1];
for(int i=0; i <= argc; i++) {
rArray[i] = argv[i];
}
// use rArray
delete [] rArray;
请注意,argv[argc]
保证为 NULL
.我已经更新了循环以复制它(因此看起来不寻常的 i<=argc
退出条件)
Note that argv[argc]
is guaranteed to be NULL
. I've updated the loop to copy this as well (hence the unusual looking i<=argc
exit condition)
如果你真的想复制字符串的内容(正如 minitech 建议的那样),代码会变得有点复杂:
If you really want to copy the content of the strings (as minitech suggests), the code becomes a bit more complicated:
rArray = new char*[argc+1];
for(int i=0; i < argc; i++) {
int len = strlen(argv[i]) + 1;
rArray[i] = new char[len];
strcpy(rArray[i], argv[i]);
}
rArray[argc] = NULL;
// use rArray
for(int i=0; i < argc; i++) {
delete [] rArray[i];
}
delete [] rArray;
这篇关于将 argv 复制到新数组中的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!