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

I get the error message in the title ? Because you are passing double variable address and catching with char* so compiler is saying

expected ‘char *’ but argument is of type ‘double *’.

So you have two ways either typecast the address of double variable as char* or catch with double* itself.

Case 1:

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

int main(int argc, char** argv) {
        double a = 1.0;
        double b = 2.0;
        printf("linea: %d - antes a(%f) b(%f)\n", __LINE__, a, b);
        swap(&a, &b, sizeof(double)); //This line gives the error
        printf("linea: %d - despues a(%f) b(%f)\n", __LINE__, a, b);
        return 0;
}

Case 2 :- Second way is typecast the address of double variable as char* send it to swap()

void swap( char *pA,  char *pB, unsigned int tam) {
        for( unsigned i = 0; i < tam; i++) {
                char tmp = *pA;
                *pA = *pB;
                *pB = tmp;
                pA = pA + 1;
                pB = pB + 1;
        }
}
int main(int argc, char** argv) {
        double a = 1.0;
        double b = 2.0;
        printf("linea: %d - antes a(%f) b(%f)\n", __LINE__, a, b);
        swap((char*)&a, (char*)&b, sizeof(double)); //typecast &a to (char*)&a
        printf("linea: %d - despues a(%f) b(%f)\n", __LINE__, a, b);
        return 0;
}