Unix中的C ++ Shell，execv：从函数动态创建并返回可用的第二个参数

Question

Everywhere I've looked for answers to this question, I see people making one small char * array of size like two and hardcoding in paths for execv. What I need to do is to take a string of parameters with the path as the first set of characters, tokenize them, and then put them in an array of char *s that execv will accept.

here is my tokenization function

char ** stringToVectorToCharArray(string inputString)
{
stringstream ss(inputString);
cout << "\n inputString is: " << inputString <<"\n";
vector<string> tokens;
tokens.clear();

while(ss >> inputString)
{
    tokens.push_back(inputString);
    }


int size = tokens.size();       
char **args = new char*[size + 1];

int i = 0;
for(; i < size; i++)
    args[i] = const_cast<char *>(tokens.at(i).c_str());

args[i + 1] = (char *) 0;

return args;
}

And this is called from

 char **args = stringToVectorToCharArray(inputString);

 execv(executeChar, args);

Within the Child section of my fork() if-else statements for flow control. I get a bad_alloc error, but I'm not sure which of my allocation statements are right, if any are for that matter. I know the return has to be in the form

 char *const argv[]

But I'm not sure how to set that up.

Answer 1

You are returning memory from a local variable ( tokens ) from your function.

for(; i < size; i++) {
  // stores pointer to local memory: args[i] = const_cast<char *>(tokens.at(i).c_str());
  args[i] = new char[tokens.at(i).size()+1];  // Create dynamically allocated copy
  strcpy(args[i], tokens.at(i).c_str());
}

The above should fix the problem. Technically, this would create a memory leak, since the memory is never deallocated, but your call to execv will replace your executable and effectively deallocate the memory.