passsing and returning char array

Bine · 02-01-2003

i have to send char array to a function and then return the char array?so what is the method for an char array of size 15?

**Azuth** · 02-01-2003

Do you want to send the contents of the array, or a pointer to it?

**newbie_grg** · 02-01-2003

i think this should do.

Code:

char* myfuntion(const char* your_ variable)
{
// do stuff
return (//your char* value here);
}
int main(void)
{
char v[] = "my array";
myfunction(v);
}

**newbie_grg** · 02-01-2003

incase of 15 elements just do

Code:

char v[15] = "my name is fala";
myfunction(v);
//same as before.

**grib** · 02-01-2003

Just use std::string. Otherwise your options are to change the string you are passed or return a dynamically allocated string and remember to dealocate it.

Code:

// with std::string, copying string
std::string sixtofour(std::string str) {
    for(int i=0;i<length;++i) {
        if(str[i] = '6') str[i] = '4';
    }
     return str;
}

char str[15] = "25 or 624";
cout << sixtofour(str) << endl;
// prints "25 or 424", str unchanged;

//with char[15]; modifying string
char * sixtofour(char str[15]) {
   int length = strlen(str);
    for(int i=0;i<length;++i) {
        if(str[i] = '6') str[i] = '4';
    }
    return str;
}

cout << sixtofour(str) << endl;
// prints "25 or 424", str="25 or 424" ;
char *s= sixtofour(str);
s[3] = '2';  // changes str!
char *s2 = sixtofour("25 or 6 to 4");// This will often crash many computers

//with char15, copying string.
char *sixtofour(char str[15]) {
    int length = strlen(str);
    char * copy = new char[length+1];
    copy[0] = '\0';
    strcat(copy,str);
    for(int i=0;i<length;++i) {
        if(copy[i] = '6') copy[i] = '4';
    }
    return copy;  
}

char *s = sixtofour(str);
//str unchanged
s[3] = 2;
//still unchanged
//but you need to do some more
delete [] s;
// clean up after sixtofour()  solo

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