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I'm pretty sure I covered this in C, but there was something about the way the ints were stored that this function wouldn't update the variables properly... Can anyone confirm, or expand?

1st Sep 2016, 6:20 PM
Nils Kaplan
Nils Kaplan - avatar
3 Respuestas
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The swap function is a classic. I believe this is what you are talking about: //arguments: two int variables void swap(int &a, int &b) { int c; c = a; a = b; b = c; } //call: swap(var1, var2) This function uses references to get a direct access to the variables, with the changes being persistent. A reference is close to a pointer as they both hold an address, but they aren't quite the same. You don't need to dereference a reference (with *) to manipulate the variable for example. Here is an equivalent with pointers: //arguments: two addresses of int variables void swap(int *a, int *b) { int c; c = *a; *a = *b; *b = c; } //call: swap(&var1, &var2) Any question?
2nd Sep 2016, 8:16 AM
Zen
Zen - avatar
+ 2
...That's a vague description. Could you find the code bit you are talking about and post it here?
1st Sep 2016, 7:57 PM
Zen
Zen - avatar
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it was the pointers section of cs50x if I remember right, I don't have the code to hand... The example had three variables, a b and c and the program was meant to swap the data from a and b around using c as an intermedeary. I don't have the code to hand though 😥
1st Sep 2016, 10:45 PM
Nils Kaplan
Nils Kaplan - avatar