I want to declare a string, initialize it by passing it by reference, and then pass it by value to the 'outputfile' function.
The code below works, but I don't know why. In main I would expect to pass the string 'filename' like
startup(&filename)
But that gives an error, and the code below doesn't. Why? Also, is there a better way to do this without using a return value?
#include <iostream>
#include <string>
using namespace std;
void startup(std::string&);
void outputfile(std::string);
int main()
{
std::string filename;
startup(filename);
outputfile(filename);
}
void startup(std::string& name)
{
cin >> name;
}
void outputfile(std::string name)
{
cout << name;
}
Your code works as expected.
&filename returns the memory address of (aka a pointer to) filename, but startup(std::string& name) wants a reference, not a pointer.
References in C++ are simply passed with the normal "pass-by-value" syntax:
startup(filename) takes filename by reference.
If you modified the startup function to take a pointer to an std::string instead:
void startup(std::string* name)
then you would pass it using the address-of operator:
startup(&filename)
As a side note, you should also make the outputfile function take its parameter by reference, because there's no need to copy the string. And since you're not modifying the parameter, you should take it as a const reference:
void outputfile(const std::string& name)
For more info, here are the rules of thumb for C++ regarding how to pass function parameters.
Related
Consider this MCVE:
#include <boost/program_options.hpp>
#include <iostream>
#include <map>
namespace po = boost::program_options;
using namespace std;
po::options_description createOptions(const std::string& description, const map<string, string>& opts) {
po::options_description newoptions(description);
for (const auto& [k, v] : opts) {
newoptions.add_options()(k, v);
}
return newoptions;
}
int main() {
map<string, string> descMap = {
{ "key", "description" },
{ "hello", "world" }
};
auto opts = createOptions("My options", descMap);
cout << opts << endl;
}
I am trying to write a convenience function to reduce the amount of C&P code when inserting similar options into an options_description object (the original code uses notifiers which were removed for simplicity, but add even more boilerplate).
To my surprise, there is no options_description_easy_init::operator() overload that accepts std::string, thus the example fails to compile.
While I could easily make the example work by calling .c_str() on k and v within the for loop, of course this would be dangerous. Is there any reason why the boost devs left out such an important overload? Why didn't they use const std::string& as argument in the first place?
And how can I make this code work without .c_str()? There is no indication that the pointer memory will be copied internally (which would be odd anyway) and I really don't feel like going back in time and managing memory on my own :-)
Looking into the implementation, it seems that internally the const char* argument passed to options_description_easy_init::operator() is wrapped by a new option_description object, which eventually converts the argument into a std::string. So as commented by #pptaszni, it is safe to call .c_str() on the std::string arguments to pass them to the program options.
What I still don't understand, however, is why there is not an std::string overload. I consider this a design flaw (also considering that options_description has a constructor taking std::string).
I would like to be able to do the following:
std::cout << str_manip("string to manipulate");
as well as
std::string str;
str_manip(str);
std::cout << str;
For this, I have two functions
#include <string>
// copying
std::string str_manip(std::string str)
{
// manipulate str
return str;
}
// in-place
void str_manip(std::string& str)
{
// manipulate str
}
but they produce the following error:
error: call of overloaded 'str_manip(std::__cxx11::string&)' is ambiguous
How can I overcome this?
The problem is with this call:
std::string str;
str_manip(str);
std::cout << str;
The compiler doesn't know which version of str_manip to call.
You can change your functions to look like this:
#include <string>
// copying
std::string str_manip(const std::string& str)
{
std::string dup = str;
// manipulate dup
return dup;
}
// in-place
void str_manip(std::string& str)
{
// manipulate str
}
Now, the compiler knows that the ambiguous call has to be the function that takes the non-const parameter. You can also be sure that your call that returns a std::string to the << operator isn't modifying your string.
This might be not the thing you are looking for, but for your code
std::cout << str_manip("string to manipulate");
the parameter to str_manip is not a string but const char* (actually an array, but convertible to a char pointer). You can overload based on that.
std::string str_manip(const char* s)
{
std::string str(s); // create str
// manipulate str
return str;
}
However, let's look at the big picture. When you see str_manip in your code, does this mean "change the string" or "make a new string based on the given string"? Do you want to be intentionally ambivalent on the real meaning?
Consider yourself reading your code in 1 year in future. What will you think when you see a call to str_manip - does this mutate its parameter? Does the answer to the previous question depend on context?
The goal in writing code is to make it clear, especially in a multi-paradigm language like C++. So, in my opinion, just don't do overloading that you are thinking about. Instead, make 2 distinct names, like
void frobnicate_str(std::string&) {...}
std::string get_frobnicated_str(std::string) {...}
Based on the idea of this entry Is it a good idea to return “ const char * ” from a function?
I thought to extend this with another question I have.
Consider the following code:
#include <string>
#include <cstdio>
const char * GetSomeString()
{
std::string somestlstring;
somestlstring = "Hello World!";
return somestlstring.c_str();
}
int main()
{
const char * tmp = GetSomeString();
printf("%s\n", tmp);
return 0;
}
If I build it with
g++ source.cpp -o executable
and execute that, I get strange symbols displayed. This is because somestlstring is destroyed through the callstack and the pointer you keep after returning became invalid.
My question is: how should I design a method or function that does not have such behaviour without actually declaring additional global variables or potential member functions?
You should drop the whole C mindset and start writing C++:
#include <string>
#include <iostream>
std::string GetSomeString()
{
std::string somestlstring;
somestlstring = "Hello World!";
return somestlstring;
}
int main()
{
std::string tmp = GetSomeString();
std::cout << tmp << std::endl;
return 0;
}
One obvious solution is to make the return type std::string.
how should I design a method or function that does not have such beahviour without actually declaring additional global variables or potential member functions?
Not at all. If you return a const char *, your function is kind of telling the caller "here you have a C string to use, but it stays mine" *), and this implies the caller doesn't have to bother releasing the resources, for example. So you can do this from an instance method (returning a pointer to a field) or you can have a function return a pointer to some static buffer (global variable).
If you want to return a dynamically allocated C string from a function, you must return char * instead and the caller has to free() it when done using it.
That all said, in C++ this doesn't make much sense, except when somehow interfacing with C code. If you want to write C++ code, go with nvoigt's answer.
*) this is thinking in terms of ownership, which is very helpful dealing with manually managed resources. The owner of something is responsible for appropriate cleanup. You can only return a const raw pointer if you don't transfer ownership of the object to the caller.
You are currently referencing the memory of a local std::string object which is destroyed when the object goes out of scope (when returning from the function)
if you really want to return a const char *:
you have to make your std::string static (but only 1 value is shared by your application)
or you have to duplicate the string memory (but you need to free it or you get memory leaks, like happened a lot with the old str() method of the old strstream object, which was later converted to std::string)
But as others said, better stick to C++ std::string (or const reference) as a return value and take c_str() of that returned string when needed for C-style interfaces.
std::string tmp = GetSomeString();
FILE *f = fopen(tmp.c_str(),"r");
The local string variable in the GetSomeString() function will get out of scope after you returned from the funtion. You will be printing random stuff that is in the memory position where the string was before. Try this:
#include <string>
#include <cstdio>
void GetSomeString(std::string& str)
{
str = "Hello World!";
}
int main()
{
std::string str;
GetSomeString(str);
std::cout << str << std::endl;
return 0;
}
I have a the following code:
#include <iostream>
using namespace std;
void func(char * aString)
{
char * tmpStr= new char[100];
cin.getline(tmpStr,100);
delete [] aString;
aString = tmpStr;
}
int main()
{
char * str= new char[100];
cin.getline(str,100);
cout<< str <<endl;
func(str);
cout<< str <<endl;
return 0;
}
Why the second cout does not print the second input string? How can I change this code to work it?
As GregS has said, the simplistic answer is to declare your function using a reference:
void func(char *&aString)
However it is not really the best solution. In C++ you generally avoid simple arrays and use containers.
#include <iostream>
#include <string>
void func(std::string &s)
{
std::getline(std::cin, s);
}
int main()
{
std::string str;
func(str);
std::cout << str << std::endl;
return 0;
}
Because the second cout will print what is pointed by str. And str, the pointer, in your main function will have the same value before and after the call to func.
Indeed, in the func function, you are changing the value of the aString variable. But this is another variable than str in main.
If you want the value of str to be changed, you have to pass it to func by reference or by pointer. (Note that what you write, is to pass the characters by pointer. I mean you have to pass the pointer by pointer: void func(char **str_ptr), or by reference void func(char *&str_ref))
If you're really doing C++, you should use std::string instead of the old C strings.
An example of passing the pointer by pointer:
func(char ** aString)
{
char * tmpStr= new char[100];
cin.getline(tmpStr,100);
delete [] *aString;
*aString = tmpStr;
}
Plus you should call it like this: func(&str);
When func() is called from main() the value of str pointer is passed to the function (this is done by copying it's value to the stack)
The value that was stored on stack when calling func() becomes a local variable aString within func(). You can modify this value but as soon as func() returns all of it's local variables will be discarded. Value of aString won't be copied back to str.
For your code to work you have to either:
Use the buffer pointed to by aString to read data: cin.getline(aString ,100);
or
Pass pointer to pointer: void func(char **aString)
Change func to
void func(char * aString)
{
cin.getline(aString,100);
}
and it works, at least it did for me.
The pointer is passed by value. Yes, you can change the content of what that pointer points to, but the old address itself is preserved when you exit the function. Hence, "aString=tmpStr"; becomes useless and "char * tmpStr= new char[100];" creates a memory leak. You need to pass the pointer by reference:
void func(char*& aString)
{
char * tmpStr= new char[100];
cin.getline(tmpStr,100);
delete [] aString;
aString = tmpStr;
}
If your aim is to actually read a line from the keyboard, do this:
std::string foo;
std::getline(std::cin, foo);
Otherwise, when you pass a pointer to a function, the pointer is passed by value. This means you cannot change the pointer itself from within the function, but you can change the object it points to. In C++ you could do this as follows:
void bar(std::string & s) {
std::getline(std::cin, s);
}
// in calling code
std::string foo;
bar(foo);
This passes a reference to the string to the function. The function can now change the contents of the string.
If you want to write a function that allocates some memory to store a result in, do it like this:
boost::shared_array<char> foo() {
boost::shared_array<char> result(new char[100]);
std::cin.getline(result.get(), 100);
return result;
}
assignment to the parameter aString within the function has no effect on str in main()
you might try
return aString
and in main
str = funct(str);
But in fact there's probably no reason to pass str into the function.
The line aString = tmpStr just changes the value of astring (i.e. an adress/pointer) to another value (i.e. another adress/pointer), but not the content of the momory pointed to by aString.
You could try to change the signature to:
void func(char ** aString)
And changing the last two lines of func to:
delete [] *aString;
*aString = tmpStr;
So the last line causes the program to change the adress stored in the memory pointed to by aString to the adress newly allocated (tmpStr). (I know, it's mind-boggling.)
And then call it via
func(&str);
You need to pass str as a reference.
#include <iostream>
void func(std::istream& is, std::string& aString)
{
std::getline(is, aString);
}
int main()
{
std::string str;
std::getline(std::cin, str);
if(std::cin)
std::cout<< str << '\n';
std::string str;
func(std::cin, str);
if(std::cin)
std::cout<< str << '\n';
return 0;
}
The pointer aString which you pass to func() indicates where (at which address) you can read and write the string str. When you then say aString = tempStr, you're replacing its original value (the address of the string pointed to by str) with a new value (the address of the string pointed to by tempStr). This will not cause str to point to the same place as tempStr.
Imagine I give you a piece of paper with my friend's home address it. Then you scratch it out and write the address of your brother (who lives in London) on it. If I then decide to visit my friend, I won't find myself traveling to London, even though you may have a piece of paper that says "Paul's friend: 123 London".
To get main to print out the string you entered in func(), you can either copy the input string to aString (you move your brother to my friend's house), or pass a pointer or reference to str (have me give you my address book, so you can change the entry for my friend to London). Either way, next time I visit "my friend's house", your brother will be there.
I am very new to C++ and still trying to learn syntax and best practices.
I've defined a method with a single parameter:
void foo(const std::string& name)
1) Is this a proper parameter declaration for a function that will be taking in a string defined by the user in, for example, a main method?
2) If this is proper/recommended syntax, what would an instantiation of a sample parameter look like?
Yes, that is the correct syntax. You can call it and provide parameters several different ways:
With a string literal:
foo("bar");
With a string variable:
std::string b = "bar";
foo(b);
With the result of a function return type string:
std::string quux();
foo(quux());
With a char* variable:
int main(int argc, char const* argv[]) {
foo(argv[0]);
}
I'm not sure if I fully understand your question, but I'll try to clarify it.
You use the terminology 'method'. I'm assuming that your method is encapsulated in a class? If so, then :-
In your header file (eg. source.h),
class dog
{
...
public:
void foo(const std::string &name);
...
};
In your source file (eg. source.cpp)
void dog::foo(const std::string &name)
{
// Do something with 'name' in here
std::string temp = name + " is OK!";
}
In your 'main' function, you can instantiate your 'dog' class, and call the 'foo' function like :-
void blah()
{
dog my_class;
my_class.foo("Testing my class");
}
If you want a function (ie. a 'method' that is not encapsulated within a class), then what you have is correct.
In your source file (eg. source.cpp)
void foo(const std::string &name)
{
// Do something with 'name' in here
std::string temp = name + " is OK!";
}
If you want to be able to call your function from outside that particular source file, you'll also need to forward declare your function in a header file.
In your header file (eg. source.h)
void foo(const std::string &name);
To call your function,
void blah()
{
foo("Testing my class");
}
Hope this helps!
1)
It is a proper parameter declaration if function foo() doesn't mean to change the string. The 'const' keyword is used to signify that the string won't be changed by the receiver.
If you write code in foo() which modifies the string you will get compiler error/warning.
2)
std::string theString = "Hello";
foo( theString );
Is this a proper parameter declaration for a function that will be taking in a string defined by the user in, for example, a main method?
Yes.
#include <string>
using namespace std;
void foo(const string& name)
1) Yes, that's a very good way to do it if you only need to read the string in the function.
2) There is no instantiation going on?
1) For most functions, it would be a fine signature. However, since you mentioned main(), there are only two valid signatures:
int main()
int main(int argc, const char* argv[])
...as you can see, you have to use C-style strings due to C-legacy compatibility (and efficiency)
2) Not sure I understand your second question, but since std::string has a constructor that takes a const char*, you can just say:
foo("hello");
or:
std::string input;
std::cout << "Enter some text: ";
std::cin >> input;
foo(input);