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here is code for counting sorting

#include <iostream>
using namespace std;

int main(){
    int a[]={2,3,1,2,3};
    int n=sizeof(int)/sizeof(int);
    int max=a[0];
    for (int i=1;i<n;i++) {
        if (a[i]>max) { 
            max=a[i];
        }
    }

    int *output=new int[n];
    for (int i=0;i<n;i++) {
        output[i]=0;
    }
    int *temp=new int[max+1];
    for (int i=0;i<max+1;i++) {
        temp[i]=0;
    }
    for (int i=0;i<n;i++){
        temp[a[i]]=temp[a[i]]+1;
    }
    for (int i=1;i<max+1;i++) {
        temp[i]=temp[i]+temp[i-1];
    }
    for (int  i=n-1;i>=0;i--) {
        output[temp[a[i]]-1]=a[i];
        temp[a[i]]=temp[a[i]]-1;
    }
    for (int i=0;i<n;i++) {
        cout<<output[i]<<"  ";
    }
    return 0;
}

but output is just 2,only one number. what is wrong i can't understand please guys help me

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3
  • What a mess. I formatted your code for you. Commented Oct 2, 2011 at 19:06
  • Why not delete the arrays on exit? Commented Aug 26, 2013 at 9:27
  • int n=sizeof(int)/sizeof(int); should be int n = sizeof(a) / sizeof(*a); Commented Jan 11, 2016 at 17:04

3 Answers 3

Reset to default
2 
int n=sizeof(int)/sizeof(int);

is wrong. That just assigns 1 to n.

You mean

int n=sizeof(a)/sizeof(int);

I've not looked beyond this. No doubt there are more problems, but this is the most significant.

This is the kind of thing you can work out very easily with a debugger.

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4 Comments

...or even with a few judiciously placed print statements.
@Greg They don't even need to be placed that judiciously!! ;-)
I think you should propose const int n=sizeof(a)/sizeof(int) instead :)
Let's learn to be safe until we know how to not to mess up (oh, and then keep the good habit anyway and profit from full optimization). It spares from analysing 50 lines of compact code to see whether something accidentally wrote to n.
1

Look at this expression:

int n=sizeof(int)/sizeof(int);

What do you think the value of n is after this? (1)
Is that the appropriate value? (no, the value should be 5)
Does that explain the output you are seeing? (yes, that explains why only one number is shown)

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2 Comments

@user466534, of greater concern is that you were unable to debug this yourself. This indicates that you are not stepping through your code in a debugger, and could not find the questions and answers that would've led you to the right answer. Everyone makes typos. Good programmers have the skills to diagnose the typos.
no ,first you are correct that good programmers have skills to diagnose the typos,just i am very tired and could not think about such simplest mistake,but thanks for advice
1

My advice would be that if you're going to do this in C++, you actually try to use what's available in C++ to do it. I'd look up std::max_element to find the largest element in the input, and use an std::vector instead of messing with dynamic allocation directly.

When you want the number of elements in an array in C++, you might consider a function template something like this:

template <class T, size_t N>
size_t num_elements(T const (&x)[N]) { 
    return N;
}

Instead of dumping everything into main, I'd also write the counting sort as a separate function (or, better, a function template, but I'll leave that alone for now).

// warning: Untested code:
void counting_sort(int *input, size_t num_elements) { 
    size_t max = *std::max_element(input, input+num_elements);

    // allocate space for the counts.
    // automatically initializes contents to 0
    std::vector<size_t> counts(max+1); 

    // do the counting sort itself.
    for (int i=0; i<num_elements; i++)
        ++counts[input[i]];  

    // write out the result.
    for (int i=0; i<counts.size(); i++)
        // also consider `std::fill_n` here.
        for (int j=0; j<counts[i]; j++)
            std::cout << i << "\t";
}

int main() { 
    int a[]={2,3,1,2,3};

    counting_sort(a, num_elements(a));
    return 0;
}
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