How do I set a pointer to Null at run-time (C++)

While you have written will set tempRow to NULL, it wont release the memory you have allocated. For that you need

free(tempRow);
tempRow = NULL;

However if you're using C++ as the tags suggest, you'd be better off using C++ new/delete

double* tempRow;
tempRow = new double[size];
   ...
delete [] tempRow;
tempRow = NULL;

you can even use the STL to handle your memory allocation for you.

std::vector<double> tempRow(size);
// You can access the data, in a similar way to an array
tempRow[5] = tempRow[4]+tempRow[3];

// If you really need to access the underlying pointer, (To pass to another 
// function for example) you can do this. Note that the pointer will be valid
// until the vector is destroyed or modified in certain ways that can cause the
// vector to reallocate its memory. So you can't use this to pass data to a 
// function that destroys or takes ownership of the passed in pointer.

fn_requiring_pointer( &temp[0] );

// The memory used in tempRow will get cleaned up automatically when the 
// object goes out of scope
//
// If I really need to free up the memory used in it early I can use a swap 
// hack. (iirc tempRow.clear() isn't guaranteed to release the memory)
std::vector<double>().swap(tempRow); // Unneeded in most cases.

Also trying to reuse the tempRow pointer for something unrelated is probably not necessary. Just create a new pointer with a different name. Reusing a variable form multiple different unrelated purposes can make code very hard to understand later.

I'm new at C++ as well, but a while ago, someone told me that using std::vector is a much safer approach to handling arrays of data.

• Automatic re-allocation when adding more elements.
• Iterators for use with stuff from #include <algorithm>.
• Bounds-protection with .at(index) element access.
• No messy pointer-tracking required.
• C-array style access with operator[].
• RAII.

You would declare a vector like this:

std::vector<double> tempRow(size);

tempRow[0] = 3.00;
tempRow[1] = 1.00;

// no need to use delete[] or free(), it will destruct itself
// and relinquish its resources automatically.

The example you've shown should work.
Also if you've not freed the memory before making temRow NULL, you are leaking memory.

double* tempRow;
tempRow = (double*) malloc(size * sizeof(double));
   ...
free(tempRow);  // free the memory.
tempRow = NULL; // reset the pointer.
   ...
tempRow = &some_other_double_var; // reuse the pointer.

Doesn't seem to work?

That's the worst complaint and a solution providers's nightmare.

Do you mean you get a compilation error?

If yes, did you include <cstdio>? and using namespace std;

Doesn't work in what way? The normal way to "unset" a pointer in C++ is with:

tempRow = 0;

but what you have should be fine, assuming you've included the correct headers or otherwise have the correct definition for NULL.

As an aside, you should first call free() on that memory before losing the pointer, otherwise you'l have a memory leak (and this is assuming you have a good reason to use C-style malloc/free instead of the more kosher C++ new/delete).