i just wanna make learn how can i write this c++ code in python ?
#include <iostream>
void modify(int* a){
*a = 20;
}
int main(){
int x = 10;
modify(&x);
std::cout << x;
// OUTPUT: 20
return 0;
}
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You can imitate that by putting the integer to be mutated in a list, but it's not idiomatic. – Passerby Dec 09 '21 at 08:49
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Python doesn't Have Pointers – Dec 09 '21 at 08:49
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i know python doesnt have but cant i make a function like this in python – cacarekt Dec 09 '21 at 08:50
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Is this helpful: https://stackoverflow.com/a/986145/9814037 ? Pretty decent answer from the owner. – Simon Provost Dec 09 '21 at 08:50
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Furthermore, About pointer here is a theory: https://stackoverflow.com/a/3106752/9814037 – Simon Provost Dec 09 '21 at 08:52
3 Answers
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This can not work for integers because integers in python are completely immutable. You can not change an integer in python. When you do for example a += 1, you create a new int with the value a + 1 and bind the variable a to it. Now, the easiest way to do this would be to have some reference type, like a class.
class IntRef:
def __init__(self, a):
self.a = a
def modify(some):
some.a = 20
ir = IntRef(10)
modify(ir)
print(ir)
DownloadPizza
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The simple answer is "you can't". But, if you had a use case where equivalent behaviour was required, the easiest way would be to wrap your variable into a mutable type. I would do the following:
def modify(a):
a[0] = 20
x = [10]
modify(x)
print(x[0])
defladamouse
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yes this works but i didnt understand why works. Shouldnt be a local variable in function? – cacarekt Dec 09 '21 at 09:00
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@cacarekt `a` indicates the _list_, not the _contents_ of the list. So when we say `a[0] = 20` we're changing the _contents_ of the list, but the list itself is unaffected. Specifically, `a` is a reference to the start of the list. We haven't changed where it starts, only what comes after. – defladamouse Dec 09 '21 at 09:04
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Abusing a list for this will be incredibly bad for performance. This isnt C, lists arent fixed size and compiled away. Use a class instead – DownloadPizza Dec 10 '21 at 05:40
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@DownloadPizza there was no indication of performance being a concern. This is simpler, a class is faster, which is better is entirely context dependent. – defladamouse Dec 10 '21 at 14:01
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Not with an int as everyone has described. The "nicest" way depends on your goal but I just use a numpy array.
def modify(a):
a += 10
x = np.array(10)
modify(x)
print(int(x))
Jono_R
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