Question

i am trying to determine what does this code?

#include <cstdlib>
#include <iostream>
#include<string.h>

using namespace std;

char *skip(char *p,int n)
{
    for (;n>0;p++)
        if (*p==0) n--;

    return p;      
}

int main(int argc, char *argv[])
{
    char *p="dedamiwa";
    int n=4;
    cout<<skip(p,n)<<endl;
}

When I run it om dev c++,it wrotes

`basic_string::copy`

When i run it on ideone.com,it wrotes

prog.cpp: In function ‘int main(int, char**)’:
prog.cpp:15: warning: deprecated conversion from string constant to ‘char*’
prog.cpp:18: warning: ignoring return value of ‘int system(const char*)’, declared with attribute warn_unused_result

Answer 1

It skips n characters of the char array.

It interpret the first parameter as a pointer to an array of character
containing at least n null characters, and return a pointer after the
n-th such null-character. Per se, there is no undefined behavior if
you pass correct input to it.

Since you pass a simple null terminated string, it has undefined
behavior, as there is only one such null-character in its input. It
will access memory after the end of the string.

Concerning the compilation errors, in C++ a constant string is of type
const char*, not char*, and you should check the return of the system
function for error. by — Sylvain Defresne

A version of the code with explicit braces is maybe a little bit
more readable for you:

using namespace std;
char *skip(char *p,int n){
      for (;n>0;p++)
        if (*p==0) {
            n--;
        }
        return p;
}

To get rid of the error:

int main(int argc, char *argv[])
{
    // cast the string which is of the type const char* to the
    // type of the defined variable(char*) will remove your warning.   
    char *p= (char*) "dedamiwa";
    int n=4;
    cout<<skip(p,n)<<endl;
}