Answer by LeeProgrammer for C++ Swapping Pointers
void swapPointer(int* &ptr1, int* &ptr2) { int* temp = ptr2; ptr2 = ptr1; ptr1 = temp;}It can be resolve by using reference.
View ArticleAnswer by jake_asks_short_questions for C++ Swapping Pointers
If you are into the dark arts of C I suggest this macro:#define PTR_SWAP(x, y) float* temp = x; x = y; y = temp;So far this has worked for me.
View ArticleAnswer by zar for C++ Swapping Pointers
The accepted answer by taocp doesn't quite swap pointers either. The following is the correct way to swap pointers.void swap(int **r, int **s){ int *pSwap = *r; *r = *s; *s = pSwap;}int main(){ int *p...
View ArticleAnswer by Alec Danyshchuk for C++ Swapping Pointers
You are not passing by reference in your example. This version passes by reference,void swap2(int &r, int &s){ int pSwap = r; r = s; s = pSwap; return;}int main(){ int p = 7; int q = 9;...
View ArticleAnswer by Euro Micelli for C++ Swapping Pointers
You passed references to your values, which are not pointers. So, the compiler creates temporary (int*)'s and passes those to the function.Think about what p and q are: they are variables, which means...
View ArticleAnswer by Ed Heal for C++ Swapping Pointers
The line r=s is setting a copy of the pointer r to the copy of the pointer s.Instead (if you do not want to use the std:swap) you need to do thisvoid swap(int *r, int *s){ int tmp = *r; *r = *s; *s =...
View ArticleAnswer by taocp for C++ Swapping Pointers
Inside your swap function, you are just changing the direction of pointers, i.e., change the objects the pointer points to (here, specifically it is the address of the objects p and q). the objects...
View ArticleC++ Swapping Pointers
I'm working on a function to swap pointers and I can't figure out why this isn't working. When I print out r and s in the swap function the values are swapped, which leads me to believe I'm...
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