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void swap(int &first, int &second){
int temp = first;
first = second;
second = temp;
}
int a=3,b=2;
swap(a,b);
The compiler complaints that void swap(int &first, int &second) has a syntax error. Why? Doesn't C support references?
334
+ A[n-1] int sum(const int* A, const int n) { int result = 0; for(int i = 0; i < n; i++) { result += A[i]; } return result; } Out-parameters. An out-parameter represents information that is passed from the function back to its caller. The function accomplishes that by storing a value into that parameter.
The out parameter in C# is used to pass arguments to methods by reference. It differs from the ref keyword in that it does not require parameter variables to be initialized before they are passed to a method. The out keyword must be explicitly declared in the method's definition as well as in the calling method.
C doesn't actually have "input" and "output" parameters, at least not directly. (Some languages do.) All arguments are passed by value, meaning that the function gets the value of the argument, but any modifications to the parameter are invisible to the caller.
The in, ref, and out Modifiersref is used to state that the parameter passed may be modified by the method. in is used to state that the parameter passed cannot be modified by the method. out is used to state that the parameter passed must be modified by the method.
C doesn't support passing by reference; that's a C++ feature. You'll have to pass pointers instead.
void swap(int *first, int *second){
int temp = *first;
*first = *second;
*second = temp;
}
int a=3,b=2;
swap(&a,&b);
C doesn't support passing by reference. So you will need to use pointers to do what you are trying to achieve:
void swap(int *first, int *second){
int temp = *first;
*first = *second;
*second = temp;
}
int a=3,b=2;
swap(&a,&b);
I do NOT recommend this: But I'll add it for completeness.
You can use a macro if your parameters have no side-effects.
#define swap(a,b){ \
int _temp = (a); \
(a) = _b; \
(b) = _temp; \
}
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