about delete char** p = new char*[5]

[q1670741824]

If at last I delete p only, positions to which 5 pointer_to_chars points lost a name to access, is it a memory leak case? How to solve such thing?
and each ptr_to_char points a string constant like *p[2] = "mon";

char** p = new char*[5];
char*[0]="sun";
char*[1]="mon";
...
delete p;

[giblit]

when ever you use new [] you should use delete []

What you are doing is deleting only the first element in the array not all the elements.

[Disch]

1) If you use new[] (with brackets), you must use delete[] (also with brackets).

2) literals should be const char*, not char*

3) char*[0]="sun"; <- this is nonsense. You probably meant this:
p[0] = "sun";

4) If you did not use new to allocate space, then you do not need to delete. Here, since you only did new[] for p.. you only need to delete[] p. You did not do new[] to allocate space for p[0], so you do not need to (and must not) delete[] p[0].

This code is perfectly acceptable:

const char** p = new const char*[5];
p[0] = "sun";
p[1] = "mon";
//...
delete[] p; // <- no memory leak 

5) You shouldn't use new unless you have to. This is even more true for new[]. I strongly recommend you use container classes like std::vector and std::string, rather than using new.

This code is less error prone and less dangerous:

std::vector< std::string > p;
p.resize(5);
p[0] = "sun";
p[1] = "mon";

// no need to delete anything ... it's all done automatically
//  no risk of memory leaks. 

[q1670741824]

@Disch,

but after delete [] p; the way to record each string constants' address is deallocated, would that be a problem? I mean there is no way to access those 5 strings, but they still there. Or they would be automatically released?

[long double main]

But you didn't allocate the memory for those string constants yourself, so you have nothing to worry about.

You only delete or delete[] what you allocated yourself with new or new[].

[q1670741824]

OIC thanks a lot, guys!