How to get the size of array of pointers made it with new double[...]
Using this is works
double arr[10];
sizeof(arr)/sizeof(arr[0]);
but this is not
double *arr;
arr=new double[10];
sizeof(arr)/sizeof(arr[0]);
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1 Answer
Initially you declared a double array of size 10.
double arr[10];
And then this statement gives number of array elements in the array. Yes right.
sizeof(arr)/sizeof(arr[0]);
In other programme this statement ....
double *arr;creates a pointer to double.
And this...
arr=new double[10];stores the starting address of array in arr. But arr is still a pointer to double.
This statement ...
sizeof(arr)/sizeof(arr[0]);is the culprit. Here sizeof(arr) returns size of the pointer to double instead of size of array. And sizeof(arr[0]) returns the size of double. Depending on your machine your size of pointer and size of double may be different. But it will print the quotient of the two values.
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