pass-by-reference

问题 How can I pass a variable by reference in scheme? An example of the functionality I want: (define foo (lambda (&x) (set! x 5))) (define y 2) (foo y) (display y) ;outputs: 5 Also, is there a way to return by reference? 回答1: See http://community.schemewiki.org/?scheme-faq-language question "Is there a way to emulate call-by-reference?". In general I think that fights against scheme's functional nature so probably there is a better way to structure the program to make it more scheme-like. 回答2:

问题 I've been successfully trying to pass a 2D-array to a function, see code below. What I don't get: Why does my "method C" (see code below) not work? What's the difference between &table[0][0] and &table ? Both of them point to the same memory address. The former works as argument passed to my function, the latter doesn't, error message: "no known conversion from 'int (*)[3][2]' to 'int *' for 1st argument void print_table(int *table, const int ROWS, const int COLUMNS) Thanks in advance! Alex

问题 I'm working on my intro C course assignment and I'm tasked with the following... Write code for a function that receives two parameters (a,and b) by value and has two more parameters (c and d) by reference. All parameters are double. From main, use scanf to get two numbers, then call the function, and then display both returned values to the output in a printf statement. The function works by assigning (a/b) to c and assigning (a*b) to d. While my knowledge is basic, I believe I understand

问题 I'm working on my intro C course assignment and I'm tasked with the following... Write code for a function that receives two parameters (a,and b) by value and has two more parameters (c and d) by reference. All parameters are double. From main, use scanf to get two numbers, then call the function, and then display both returned values to the output in a printf statement. The function works by assigning (a/b) to c and assigning (a*b) to d. While my knowledge is basic, I believe I understand

问题 I'm trying to create a method using mono where a string is passed by reference, here is the test code I have: C++: static bool p_TestMethod(int num, MonoString ** response) { auto b = mono_string_new(mono_domain_get(), "Test repsonse"); response = &b; return true; } //... mono_add_internal_call("SharpCode.TestClass::Testmethod", p_TestMethod); C#: [MethodImpl(MethodImplOptions.InternalCall)] public static extern bool Testmethod(int num, out string response); public bool RunTheTest() { string

问题 As much as I enjoy C++ programming, there is one thing I really don't get. To me, it seems that the most common way of programming a function is like this: some_function(a variable) do something according to the data in the variable example: bool match_name(const std::string& name) { return this->name == name; } I find myself using const ref for 90% of all function parameters in my code (maybe I'm doing something wrong). My question is: Why is a copy of the variable the "default" type of

问题 As much as I enjoy C++ programming, there is one thing I really don't get. To me, it seems that the most common way of programming a function is like this: some_function(a variable) do something according to the data in the variable example: bool match_name(const std::string& name) { return this->name == name; } I find myself using const ref for 90% of all function parameters in my code (maybe I'm doing something wrong). My question is: Why is a copy of the variable the "default" type of

问题 In C++ where we use pass by reference we reference the address of whatever it is that we passed from the argument to the parameter of the function which is essentially a pointer right? So while they are essentially the same thing, alias and all, doesnt a pointer require memory space as well? So shouldnt whatever we have in a parameter function let us call B point to the memory location of whatever the argument was that was passed let us call A which in turn is the memory location of our value

问题 In C++ where we use pass by reference we reference the address of whatever it is that we passed from the argument to the parameter of the function which is essentially a pointer right? So while they are essentially the same thing, alias and all, doesnt a pointer require memory space as well? So shouldnt whatever we have in a parameter function let us call B point to the memory location of whatever the argument was that was passed let us call A which in turn is the memory location of our value

问题 #include<stdio.h> void swap(int *a,int *b){ int p=*b; *b=*a; *a=p; /*int *p=b; b=a; a=p; */ } int main(){ int a,b; scanf("%d %d",&a,&b); swap(&a,&b); printf("%d %d",a,b); } Above is the code. If I put 3 5 as an input, then it should swap its values, and 5 3 should come out as an output. I got my answer by trying int p=*b thing However I also tried commented part, but it didn't work. So, I checked their address in swap and in main. In swap int *a and int *b their address changed However, when