命令行参数中出现字母

Question

I'm having some difficulty with the argv pointer. Using *argv[1] gives me the argc count and *argv[2] the entire argument. Shouldn't each element, when dereferenced give one character?

Thanks

int main(int argc,char *argv[])
{
    char store[10]={0,0,0,0,0,0,0,0,0,0};int freq[10]={0,0,0,0,0,0,0,0,0,0};int flag=0;int count=0;int t;
    for(int i=0,j=0;i<argc;i++)
    {
        for(int z=0;z<count;z++)
        {   
        if(i==0) break;
            if(*argv[i+2]==store[z]) flag=1;t=z;break;
        }
           if(flag==0||i==0) {store[j]=*argv[i+2];j++; count++;}
           else freq[t]+=1;
           flag=0;
    }
    for(int x=0;x<count;x++) cout<<store[x]<<"\t"<<freq[x]<<endl;
}

Answer 1

You getting the segfault at *argv[i+2] when i reaches argc-2 because then you would access the argument string argv[argc] which is not defined.

If you want to iterate through the characters of an argument, you should use a two-dimensional access, eg:

// assume there are two arguments present (argc == 3)
int len = 5; // assume there a 5 chars in argv[2]
for(int k = 0; k < len; k++) {
  // do something for each char
  if(argv[2][k] == 'x') { ... }
}

EDIT: changed example to argv[2] holding the string.