noob question how toDeclare a variable.

[donaldfarkas]

#include <stdio.h>

main()
{
	int x,y,b;
	x = 15;
	y = 10;
	b = x*y;
	printf(b);
}

how to display ? would i use printf?
would i use return b;?

and does any one recommend a good book for linux C++ programming what ive noticed is that C++ windows its a bit different like the #include <iostream> is non existent in linux or am i missing something.

[codewizzrd]

a.) Ask questions like this in the linux programming section. People here will generally get upset with questions that are about programming and not linux but....... since your asking about programming in linux.

use this:
#include <stdio.h>

main(int argc, char** argv) // Don't worry about the meaning of this yet
{
int x,y,b;
x = 15;
y = 10;
b = x*y;
printf("%d\n", b);

return 0; // Don't worry about the meaning of this yet either
}

this is a c program by the way but Im sure you already knew that.

b.) Get the book "C++ How to Program" by Deitel and Deitel. Its expensive but if you are serious its a minimal investment.

[Rubberman]

This is C code, not C++. Try this:

#include <iostream>
int main(void)
{
    int x, y, b;
    x = 15;
    y = 10;
    b = x*y;
    cout << b << endl;
    return 0;
}

Note that with C++ and newer versions of C, main() must return an integer. Also, not declaring a return value for a function (including main()) implies an integer return value.

[donaldfarkas]

This is C code, not C++. Try this:

#include <iostream>
int main(void)
{
    int x, y, b;
    x = 15;
    y = 10;
    b = x*y;
    cout << b << endl;
    return 0;
}

Note that with C++ and newer versions of C, main() must return an integer. Also, not declaring a return value for a function (including main()) implies an integer return value.

when i type in the above code it wotn work i tried this already.

[Donald@localhost ~]$ cd Desktop/C++
[Donald@localhost C++]$ make test
cc     test.c   -o test
test.c:1:20: error: iostream: No such file or directory
test.c: In function ‘main’:
test.c:9: error: ‘cout’ undeclared (first use in this function)
test.c:9: error: (Each undeclared identifier is reported only once
test.c:9: error: for each function it appears in.)
test.c:9: error: ‘end1’ undeclared (first use in this function)
make: *** [test] Error 1

[donaldfarkas]

a.) Ask questions like this in the linux programming section. People here will generally get upset with questions that are about programming and not linux but....... since your asking about programming in linux.

use this:
#include <stdio.h>

main(int argc, char** argv) // Don't worry about the meaning of this yet
{
int x,y,b;
x = 15;
y = 10;
b = x*y;
printf("%d\n", b);

return 0; // Don't worry about the meaning of this yet either
}

this is a c program by the way but Im sure you already knew that.

b.) Get the book "C++ How to Program" by Deitel and Deitel. Its expensive but if you are serious its a minimal investment.

#include <stdio.h>

main(int argc, char** argv) // On this line what those this mean? int argc, char** argv
{
	int x,y,b;
	x = 15;
	y = 10;
	b = x*y;
	printf("%d\n", b);  // what does %d mean?

        return 0;   
}

[Rubberman]

It seems you don't have the c++ compiler installed, or you need to rename your source file from test.c to test.cpp so make will know to use the c++ compiler. The C compiler doesn't know about C++ stuff like <iostream>, etc.

As for your question about the arguments passed to main(), "int argc" and "char** argv", these are passed to your program when you have arguments on the command line. If your program doesn't take any arguments, then you can use the void argument instead. Also, it is better to use "const char** argv" since you probably don't want to be able to modify the contents of the arguments directly. So, argc is the number of arguments passed to the program from the command line, including the command itself, and argv is an array of strings that contains the actual arguments themselves. So, if you executed the command "test a b c", argc would equal 4 and argv would be an array that looks like this:

const char* argv[4] = { "test", "a", "b", "c" };

As a result, you can process your arguments like this:

for (int i = 1; i < argc; i++)
{
cout << "arg[" << i << "] == " << argv[i] << endl;
}

[donaldfarkas]

changed the test file to test.cpp

[Donald@localhost ~]$ cd Desktop/C++
[Donald@localhost C++]$ make test
g++ test.cpp -o test
test.cpp: In function ‘int main()’:
test.cpp:9: error: ‘cout’ was not declared in this scope
test.cpp:9: error: ‘end1’ was not declared in this scope
make: *** [test] Error 1

this is ther error iget when i try to compile

[Rubberman]

1. It is endl (with a letter ell) not end1 (number 1).
2. It may may be that you need to include the standard namespace, so try this:

#include <iostream>

using namespace std;

int main( int argc, const char** argv)
.
.
.

[donaldfarkas]

yea i had number 1 there and also i didnt have using namespsace std;

[donaldfarkas]

b.) Get the book "C++ How to Program" by Deitel and Deitel. Its expensive but if you are serious its a minimal investment.

i managed to land a digital copy without denting a big whole in my wallet, anyway i got the book and im gonna start working on it the problem is the tutorials are in chm format so i gotta find a way linux can read it the the tutorial program files are cpp so no problem there


found a chm viewer