error: cannot convert 'double*' to 'char*' for argument '1' to 'void swap(char*, char*, unsigned int)'

Question

#include 

void swap( char* pA,  char* pB, unsigned tam) {
    for( unsigned i = 0; i < tam; i++) {
        char tmp = *pA;
        *pA = *pB;
                 


        

Answer 1

Well, your function is expecting a pointer to a character for the first two arguments

void swap( char* pA,  char* pB, unsigned tam) 

but you are passing in pointers to double

double a = 1.0;
double b = 2.0;
swap(&a, &b, sizeof(double)); //This line gives the error

The following would allow you to swap two doubles, unless there is a specific reason you are swapping one byte at a time:

void swap(double *pA, double *pB, unsigned int tam) {
    double tmp = *pA;
    *pA = *pB;
    *pB = tmp;
}