how should i declare the function if i want it to return a value like
char text[10];
i have tried this but it does not work and i dont get what im doing wrong...
char functiontest()
{
char text[10];
//i do put a value in there but lets keep it simple in here
return (text);
}
arghhhhh what is wrong?
text is a local variable to functiontest()... when the function returns, text is out of scope and is destroyed... that's why you can return it. can't return a string by value.. you have to use malloc or something like that to put it on the heap so youcan passit back...
there are a few ways to do it.. so read some tutorials and you'll figure it out.
Quidquid latine dictum sit, altum sonatur.
Whatever is said in Latin sounds profound.
I'm not sure about this one, but you could do something like this:
Code:typedef char[10] mystring; mystring myfunc() { mystring anotherstring; return anotherstring; }
MagosX.com
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Teach a man to fish and you feed him for a lifetime.
I always thought u could return an array. Does this work. I do not have a compiler here.
char [] function()
{return char array[];}
You can't return an array from a function unless you either return a pointer to the array, such as char * or wrap the array in a struct.
-PreludeCode:struct Array{ int myArray[20]; }; struct Array myFunc(int someArray[]);
My best code is written with the delete key.
Nope. You would need to do the pointer thingy and use a variable declared on the heap, not local.
Let's say you wanted to change a string in a function. The options are to pass the string to the function as a pointer/array
int main()
{
char dummy[10] = "now";
foo(dummy);
cout << dummy;
.
.
.
return 0;
}
void foo(char * _dummy) //or void foo(char [] _dummy)
{
strcpy(_dummy, "later");
}
or to declare the return string using new (or malloc) in the function itself.
int main()
{
char dummy[10] = "now";
strcpy(dummy, bar());
cout << dummy;
.
.
delete [] dummy2;//delete dummy2 heere to free memory declared in bar().
.
.
return 0;
}
char * bar()
{
char * dummy2 = new char[10];
strcpy(dummy2, "later");
return dummy2;
}
I made a test on this. You can't return the array directly, but if you store it into a struct you can (like Prelude said). However, if you're going to use this array as a string (char text[10]... hm) it is better to return a pointer.
Code:#include <conio.h> #include <iostream.h> //Structure containing an array typedef struct { char Nr[10]; }CharArray; //Function that return the array-structure CharArray MyFunc() { CharArray MyTempArray; for(int i=0; i<10; i++) MyTempArray.Nr[i]=i; return MyTempArray; } //Main function int main() { CharArray MyArray=MyFunc(); for(int i=0; i<10; i++) cout << (int)MyArray.Nr[i] << " "; getch(); return 0; }
Last edited by Magos; 01-24-2002 at 10:40 AM.
MagosX.com
Give a man a fish and you feed him for a day.
Teach a man to fish and you feed him for a lifetime.
