CODEWITHSUNDEEP
c++arrayspointers
c70u
c70u

Reputation: 119

Dynamic memory allocation and pointer non-char array type

I read this C++ tutorial, and in the pointers section there is a confusing example, exactly this:

double (*pVal2)[2]= new double[2][2]; //this will add 2x2 memory blocks to type double pointer
*(*(pVal2+0)+0) = 10;
*(*(pVal2+0)+1) = 10;
*(*(pVal2+0)+2) = 10;
*(*(pVal2+0)+3) = 10;
*(*(pVal2+0)+4) = 10;
*(*(pVal2+1)+0) = 10;
*(*(pVal2+1)+1) = 10;
*(*(pVal2+1)+2) = 10;
*(*(pVal2+1)+3) = 10;
*(*(pVal2+1)+4) = 10;

Is int (*pVal)[2] an array pointer? I do not understand why is it allocating memory for double[2][2] but the *(*pVal2+1)+4) goes to 4?

Upvotes: 0

Views: 43

Answers (1)

Barry
Barry

Reputation: 302851

Using the spiral rule:

        +--------+
        | +---+  |
        | ^   |  |
double (*pVal2)[2];
 ^      ^     |  |
 |      +-----+  |
 +---------------+

pVal2 is a pointer to an array of 2 doubles. Or, simpler:

using T = double[2];
T *pVal2 = new T[2];

The rest of the code leads to undefined behavior as: *(p + idx) is equivalent to p[idx], so *(*(pVal2+1)+4) is equivalent to pVal2[1][4].But the type of pVal2[1] is double[2], so there is no 5th element there...

Upvotes: 2

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