Array and Function issue

**xdrunkcowx** · 06-04-2003

hello i put this code in the compiler (Dev-C++ 4.0):

PHP Code:


#include <iostream>

using namespace std;

int main()

{

    int i;

    int x;

    int a;

    char square[25];

    square[0] = 'a';

    square[1] = 'b';

    square[2] = 'c';

    square[3] = 'd';

    square[4] = 'e';

    square[5] = 'f';

    square[6] = 'g';

    square[7] = 'h';

    square[8] = 'i';

    square[9] = 'j';

    square[10] = 'k';

    square[11] = 'l';

    square[12] = 'm';

    square[13] = 'n';

    square[14] = 'o';

    square[15] = 'p';

    square[16] = 'q';

    square[17] = 'r';

    square[18] = 's';

    square[19] = 't';

    square[20] = 'u';

    square[21] = 'v';

    square[22] = 'w';

    square[23] = 'x';

    square[24] = 'y';

    square[25] = 'z';

    void read () {

    for (i=0; i<=25; i++){

    cout<<square[i];

    }

    }

    for (a=0; a<=25; a++){

    for (x=0; x<=a; x++){

    read();

    }

    cout<<"\n";

    }

}

and i get a **** load of errors like:
35 test22.cpp
parse error before `{'
36 test22.cpp
parse error before `)'

then i just get a lot of sntax errors like
40 test22.cpp
syntax error before `<='

if i take out the read fuction and the calling of the read function it all compiles fine.. could someone please help me figure out what im doing rong.. thanks,
Jordan

**7stud** · 06-04-2003

You don't define functions inside other functions. You need a structure like this:

Code:

void my_func(int a)
{
    cout<<a<<endl;
}

int main()
{
    int a;
    func(a);
    
    return 0;
}

Also, you can assign the letters of the alphabet to the char array in a loop. A char type is really an int, and all characters are converted to integer codes for storage, then they are converted back to characters for display. You can do directly what the compiler is doing indirectly. If you look in an ASCII table, you will see the integer codes for a-z are 97-122, so you can do this:

Code:

for (int i=0; i<26; i++)
   square[i]= i + 97;

And, when you are ready to display the char's, it doesn't matter how the integer code got in the variable, the compiler just converts the code to a character and displays it.

**xdrunkcowx** · 06-04-2003

PHP Code:


#include <iostream>

using namespace std;

    int x;

    int a;

    int d;

    int square[25];

    for (int c=0; c<26; c++)

     square[c]= c + 97;

   }



    void read () {

    for (int i=0; i<=25; i++){

    cout<<square[i];

    }

int main()

{

    for (a=0; a<=25; a++){

    for (x=0; x<=a; x++){

    read();

    }

    cout <<"\n";

    }

}

i still get errors.. now all i want this to do is make var that changes in a for statement and that goes like this

a
next i increasment or next step or w/e
b
then
c
untill u finish the alphabet then u go
aa
ab
ac
ad
ae
untill thats done then
ba
bb
bc
bd
be
bf
untill you get something like
zzzzzzzzz

thanks,
jordan

**JaWiB** · 06-04-2003

int x;
int a;
int d;

get rid of the x and a and declare them inside your "for" loop:

Code:

for (int a=0; a<=25; a++){
    for (int x=0; x<=a; x++){
    read();
    }
    cout <<"\n";
    }

don't know what "d" is for

also:

int square[25];
for (int c=0; c<26; c++)
square[c]= c + 97;
}

move to inside of main

there might still be problems, im not looking that closely

**xdrunkcowx** · 06-04-2003

PHP Code:


#include <iostream>

using namespace std;



    void read () {

    for (int i=0; i<=25; i++){

    cout<<square[i];

    }

int main()

{

    int square[25];

    for (int c=0; c<26; c++)

    square[c]= c + 97;

    }

for (int a=0; a<=25; a++){

    for (int x=0; x<=a; x++){

    read();

    }

    cout <<"\n";

    }



}

still a billion errors.. AHh this is confusing..

thanks a lot,
jrodan

**Zach L.** · 06-04-2003

You've got some problems matching up brackets there. read() is missing a closing bracket, and main() has a closing bracket too soon.

Also, read() cannot see the variable 'square', so you'd have to pass it as an argument.

**7stud** · 06-04-2003

Anticipating the next question:

A function has a return type and parameter types:

<return type> read(<parameter list>)
{
}
If you don't need to return a value then the type is void, same for the parameter types.

Code:

void read(void)
{
   cout<<"hello";
}

int main()
{
    read();
  
    return 0;
}

Your function doesn't need to return anything to main()--it just needs to display the array--so your return type will be void.

Now for parameters. Your function needs the array to print it. To indicate an array is going to be passed to your function, you have to indicate the type in the parameter list for your function:

Code:

void read(char square[]) //The array size is not part of the type
{
    ....
}

int main()
{
   char square[26];
   
   ...

   read(square); //Passes the array square to read()
  
   return 0;
}

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