If statement debug in the program .

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Here ,
In this program when I debug , and when the control reaches the if statement which contains // symbol beside it,
At that point when I click stepover option , the control is reaching the end of the function.
Why is it happening?

C++

#include <stdio.h>
#include<stdlib.h>

void func(int i,int *nums,int j,int numsSize, int *p)
{
    if(i==(numsSize-1))
    {
        (*p)=1;
        return ;
    }
    else
    {
        int k=nums[i]+i;
        if(k>=numsSize)     //
        {
            k=numsSize-1;
        }
        for( ;(k>i) && ((nums[k]!=0)||(nums[k]==0 && k==numsSize-1)) ; k--)
        {
            func(k,nums,j,numsSize,p);
            if((*p)==1)
            {
                return ;
            }
        }
    }
}

int canJump(int* nums, int numsSize){
    int i=0;
    int j=0;
    int *p=(int*)malloc(sizeof(int));
    *p=0;
    func(i,nums,j,numsSize,p);
    if((*p)==0)
    {
        return 0;
    }
    else
    {
        return 1;
    }
}

int main()
{
    int nums[]={2,5,0,0};
    int numsSize=4;
    canJump( nums, numsSize);
}

What I have tried:

I tried searching in google...............................................................

Posted 3-Sep-21 3:09am

_-_-_-me

Updated 3-Sep-21 4:07am

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Comments

Richard MacCutchan 3-Sep-21 9:27am

Instead of wasting time with Google, look at all the values of the variables in the following for statement to see why it does not get executed.

The Other John Ingram 3-Sep-21 10:08am

put a printf statement just before the test and see just what the two variables are.

KarstenK 3-Sep-21 14:10pm

Thats why it is best practice to only write one command at one line ...

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Richard MacCutchan · Answer 1 · 2021-09-03T04:04:00

Solution 1

When I ran your code I printed the values of all the variables in the for statement thus:

k: 2 > i: 0) && ((nums[k]!=0: 0)||(nums[k]==0 && k==numsSize-1: 4

which shows:

k: 2 
i: 0
nums[k]: 0
numsSize: 4

which means the first attempt to run the statement fails because the for expression is false. Hence the function returns immediately.

Posted 3-Sep-21 4:04am

Richard MacCutchan

Comments

CPallini 3-Sep-21 10:08am

Indeed.
My 5.

CPallini · Answer 2 · 2021-09-03T04:07:00

Solution 3

It jumps at the end of the function because, with

C

k = 2; 
nums[k] = 0; 
numsSize = 4;

the condition of the for loop is false (0).

Posted 3-Sep-21 4:07am

CPallini

Comments

Richard MacCutchan 3-Sep-21 10:35am

I expect you worked it out quicker than me.

CPallini 3-Sep-21 11:45am

No way. You was faster than I.

OriginalGriff · Answer 3 · 2021-09-03T04:05:00

We can't tell: it requires you code running with your data to find out - and we have no access to either.

So start by doing a fresh full rebuild compilation (this ensures the code you look at and the EXE file are the same).
Then run your code in the debugger with a breakpoint on the if line, and the first line of code. Before you step it, look at the values of k, i, and nums[k]and work out what should happen with the if
Step it and see what did happen.

We can't do any of that for you!

If statement debug in the program .

3 solutions

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