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C# get the value when first time the array element is null

Time:09-21

get when first time element value is null I want to add the value at that position

a[0]=0
a[1]=1
a[2]=2
a[3]=3
a[4]=null --- add 4 here                                                          
a[5]=null
a[6]=null                                                                                                                 
                                                                                                                                                                                                                    

CodePudding user response:

try this, it was tested

var i=0;
    
    for ( i=0;  i<a.Length; i  )
    {
        if (a[i]==null) break;
    }
      if(i<a.Length) a[i]=4; 

CodePudding user response:

Loop through the array, check values. if you encounter null, do the thing.

for(int i = 0; i< a.Length; i  )
{
    if(a[i] == null)
    {
        // do something
        
        // use break if you dont want to continue
        break;

    }
}

CodePudding user response:

It is necessary to check with the comparison operator. "your_list?=null"

CodePudding user response:

Using a foreach would not be a bad idea:

foreach (var element in a)
{
    if (element is null) break; // or do something with it
}

CodePudding user response:

Loop through the array a and check for every iteration if the value in the array is null.

Example:

for(int i; i < a.Length; i  ) {
    if (a[i] is null) {
        //Do your work here
    }
}

CodePudding user response:

It is not C# related question actually. It is more about algorithms.

If talking about C#, first of all, if you are going to compare with Nulls, that means that you need to use Nullable type (e.g. int?[] in your case).

The solution is: You need to loop from the 0th element until nth-1 and at each step to compare value with Null. If Null item is at ith index, then replace a[i] with a new value and exit the cycle.

You can loop using for, foreach, or different Linq methods.

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