help in array

[neha007]

hi frnds i have a prob in a question actually in this question iwant to pass part of an array to a function but i cant understand how can i pass my atempt is

#include<stdio.h>
#include<conio.h>
void part(int(*f)(int a[5]));
main()
{
clrscr();
int i;
int a[5]={1,2,3,4,5};
{
for(i=0;i<5;i++)
printf("%d",a[i]);
}
void part(int(*f)(int a[i]));
getch();
}
void part(int(*f)(int a[5]))
{
int(*a[])()={
	     for(i=2;i<=3;a++)
	     printf("%d",a[i]);
	     }

tell me how can i solve my problem

[Salem]

int a[5]={1,2,3,4,5};    // this is your array

void part( int a[5] );   // this is the prototype

part( a );  // this is a call to the function - note, no [ ]

void part( int a[5] ) {  // this is the definition
  int i;
  for(i=0;i<5;i++)
    printf("%d",a[i]);
}

[neha007]

hey i know but i have to pass just a part of array not the whole array

[manutd]

You should indent correctly.

#include<stdio.h>

Ewwww.

#include <cstdio>

[laserlight]

You should indent correctly.

To be fair, this is a matter of style, and except from the change from <stdio.h> to <cstdio>, it does not matter at all.

Consider a sample program probably written by Bjarne Stroustrup: How do I write this very simple program?

hey i know but i have to pass just a part of array not the whole array

Pass two pointers: one points to the start of the range, the other to the end (or one past the end, to follow a general idiom of the C++ standard library). Alternatively, you can pass the array, and two indices: one to the start of the range, and the other to the end (you get the idea).

[manutd]

In Bjarne's case, there is actual indentation. Here is an ugly mess. I'll highlight in red everything that is wrong:

#include<stdio.h>     //See my post. Also, should be <iostream>.
#include<conio.h>     //Nonstandard, consider replacing.
void part(int(*f)(int a[5]));     //See Salem's post.

main()     //SHould be int main().
{
     clrscr();
     int i;
     int a[5]={1,2,3,4,5};
     {     //Move one line down.
     for(i=0;i<5;i++)
          printf("%d",a[i]);     //Why do this in C++? As I said, use <iostream>
     }
     void part(int(*f)(int a[i]));     //Again
     getch();
}

void part(int(*f)(int a[5]))
{
     int(*a[])()=     //No clue what you're trying to do, so I'll guess.
     {            
          for(i=2;i<=3;a++)
	          printf("%d",a[i]);
	     }     //Again, see Salem's post.
}     //You need this closing brace.
       //For all of this, you need indentation, which I've added.

[Salem]

> hey i know but i have to pass just a part of array not the whole array
Well
part( a );

Is the same as
part( &a[0] );

There is no such thing as part of an array, just a pointer to some element of the array, usually the first.

[neha007]

hey no one of u can understand wat em i talking bout actually look actually array is e.g

a{i]={0,1,2,3,4,5,6,7,8,9}

now i have to right a program in which i have to pass for example just

a[i]={3,4,5}

not the whole elements of actually array

[laserlight]

hey no one of u can understand wat em i talking bout

I have the general idea, but passing a "sub-array", if you will, is not quite right. Rather, you should pass pointers to a range in the array.

Consider the approach demonstrated in this program:

#include <iostream>

void print(int* begin, int* end);

int main()
{
    int[] numbers = {0,1,2,3,4,5,6,7,8,9};
    // Pass pointers to the array to denote a range.
    // (numbers + 3) is the start of the range {3,4,5}
    // (numbers + 6) is one past the end of the range
    print(numbers + 3, numbers + 6);
}

// Prints range denoted by [begin, end).
void print(int* begin, int* end)
{
    for (int* i = begin; i != end; ++i)
    {
        std::cout << *i << " ";
    }
    std::cout << std::endl;
}