如何将 const float* const 复制到 float*？

Question

So in my function, I was passed a parameter const float* const d_inputArray How do I read its content and copy to a different array float* d_tempInput ? I don't know how do deal with either const.

Answer 1

假设两个指针都指向相同大小的缓冲区：

memcpy(d_tempInput, d_inputArray, LENGTH_OF_BUFFER_IN_BYTES);

Answer 2

/* malloc memory for d_tempInput */
d_tempInput = (float *) malloc(sizeof(d_inputArray)); 

/* memcpy everything from d_inputArray into d_tempInput */
memcpy(d_tempInput, d_inputArray, sizeof(d_inputArray)); 

/* Free when done */
free(d_tempInput);

Answer 3

Example: const float * -> float * -> const float *

The keyword const refers to the content of the memory region the pointer points to.

Do not confuse this with a constant pointer (construction by float * const pntr ). Read Constant Pointers and Pointers to Constants for more info.

Code

Interactive version: here

#include <iostream>
#include <cstring>

int main()
{
    size_t N = 3;
    
    const float a[N] = {100, 101, 102};
    
    //float *b = a;  // not possible
    float *b = new float[N];
    std::memcpy(b, a, N*sizeof(float));
    
    const float *c = b;
    
    for (size_t ii = 0; ii < N; ++ii)
    {
        std::cout << a[ii] << "; " << b[ii] << "; " << c[ii] << std::endl;
    }
    delete b;
    return 0;
}

Output

100; 100; 100
101; 101; 101
102; 102; 102