help needed please!

[Buck Rogers]

Hi guys!

Task 1: Write a program which presents a menu with 5 options. The 5th
option quits
the program. Each option should execute a system command using system().

Below is my humble offering.
=============== =============== =============== ======
#include <stdio.h>
#include <stdlib.h>

int menu( void );

int main( void )
{
int option;

while(1) {
option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
exit(0);
default:
puts("Not a valid option, try again");
}
}
return 0;
}

int menu( void )
{
int option;

puts("1. List directory");
puts("2. help");
puts("3. time");
puts("4. type menu.c");
puts("5. Quit");
scanf("%d", &option);

return option;
}
=============== =============== =============== =============
Problem: My compiler is giving me an "unreachabl e code" warning for line
32(return 0;). The
program runs fine. Why does the compiler(Borlan d C++ Builder 6) give this
warning?

Task 2: Create a calculator program.

Below is my humble offering.
=============== =============== =============== =============== =======
#include <stdio.h>

void add(int a, int b);
void subtract(int a, int b);
void multiply(int a, int b);
void divide(int a, int b);

int main( void )
{
int a, b;
char c;

puts("Enter an integer");
scanf("%d", &a);
puts("Enter an operator");
scanf("%c", &c);
puts("Enter another integer");
scanf("%d", &b);
switch( c ) {
case '*':
multiply(a, b);
break;
case '+':
add(a, b);
break;
case '-':
subtract(a, b);
break;
case '/':
divide(a, b);
break;
default:
puts("Operator is not valid");
}
return 0;
}

void add(int a, int b)
{
printf("Sum is: %d", a+b);
}

void subtract(int a, int b)
{
printf("Answer is: %d", a-b);
}

void multiply(int a, int b)
{
printf("Product is: %d", a*b);
}

void divide(int a, int b)
{
printf("Answer is: %d", a/b);
}
=============== =============== =============== ========
Output:

Enter an integer
2
Enter an operator
Enter another integer

Problem: Why does the program not allow the user to input the operator
before
jumping straight to "Enter another integer"? I think I am missing
something very
fundamental here.

Thanks in advance guys! Love your work!

Buck
--
Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/

[Derk Gwen]

# while(1) {
# option = menu();
# switch(option) {
# }
# }
# return 0;

# Problem: My compiler is giving me an "unreachabl e code" warning for line
# 32(return 0;). The
# program runs fine. Why does the compiler(Borlan d C++ Builder 6) give this
# warning?

Because it's true? The while condition is always true, and loop is devoid
of breaks or gotos out of loop. The only one to exit the loop is to exit
the program; whether it loops forever or exits the program, no code after
the loop can be reached and executed.

--
Derk Gwen http://derkgwen.250free.com/html/index.html
So basically, you just trace.

[nrk]

Buck Rogers wrote:

Hi guys!

Task 1: Write a program which presents a menu with 5 options. The 5th
option quits
the program. Each option should execute a system command using system().

Below is my humble offering.
=============== =============== =============== ======
#include <stdio.h>
#include <stdlib.h>

int menu( void );

int main( void )
{
int option;

while(1) {
option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
exit(0);
default:
puts("Not a valid option, try again");
}
}
return 0;
}

int menu( void )
{
int option;

puts("1. List directory");
puts("2. help");
puts("3. time");
puts("4. type menu.c");
puts("5. Quit");
scanf("%d", &option);

return option;
}
=============== =============== =============== =============
Problem: My compiler is giving me an "unreachabl e code" warning for line
32(return 0;). The
program runs fine. Why does the compiler(Borlan d C++ Builder 6) give this
warning?

The only way out of your while ( 1 ) loop is through the exit(0) call, and
that doesn't bring you to the return 0; statement either. Hence the
complaint. Ideally, you should be able to replace the exit(0) call with
return 0; get rid of the existing return statements and the compiler should
be happy.
Task 2: Create a calculator program.

Below is my humble offering.
=============== =============== =============== =============== =======
#include <stdio.h>

void add(int a, int b);
void subtract(int a, int b);
void multiply(int a, int b);
void divide(int a, int b);

int main( void )
{
int a, b;
char c;

puts("Enter an integer");
scanf("%d", &a);
Read the fine FAQ. Specifically Q12.18:
http://www.eskimo.com/~scs/C-faq/q12.18.html

One fix is:
int ret;
/* read an int and discard rest of line */
ret = scanf("%d%*[^\n]", &a);
if ( ret < 1 ) {
/* ret == 0 => invalid input */
/* ret == EOF => end of input or error */
/* handle appropriately */
}
/* read and discard newline */
if ( !feof(stdin) ) getchar();
puts("Enter an operator");
scanf("%c", &c);
Similar problem as above. Similar fix can be applied.
puts("Enter another integer");
scanf("%d", &b);
switch( c ) {
case '*':
multiply(a, b);
break;
case '+':
add(a, b);
break;
case '-':
subtract(a, b);
break;
case '/':
divide(a, b);
break;
default:
puts("Operator is not valid");
}
return 0;
}

void add(int a, int b)
{
printf("Sum is: %d", a+b);
}

void subtract(int a, int b)
{
printf("Answer is: %d", a-b);
}

void multiply(int a, int b)
{
printf("Product is: %d", a*b);
}

void divide(int a, int b)
{
printf("Answer is: %d", a/b);
}
=============== =============== =============== ========
Output:

Enter an integer
2
Enter an operator
Enter another integer

Problem: Why does the program not allow the user to input the operator
before
jumping straight to "Enter another integer"? I think I am missing
something very
fundamental here.

What you're missing is knowledge of how scanf works. A trip to the scanf
pages in your C book could come in handy.

-nrk.
Thanks in advance guys! Love your work!

Buck

--
Remove devnull for email

[Al Bowers]

Buck Rogers wrote:

Hi guys!

Task 1: Write a program which presents a menu with 5 options. The 5th
option quits
the program. Each option should execute a system command using system().

Below is my humble offering.
=============== =============== =============== ======
#include <stdio.h>
#include <stdlib.h>

int menu( void );

int main( void )
{
int option;

while(1) {
option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
exit(0);
default:
puts("Not a valid option, try again");
}
}
return 0;
}

int menu( void )
{
int option;

puts("1. List directory");
puts("2. help");
puts("3. time");
puts("4. type menu.c");
puts("5. Quit");
scanf("%d", &option);

return option;
}
=============== =============== =============== =============
Problem: My compiler is giving me an "unreachabl e code" warning for line
32(return 0;). The
program runs fine. Why does the compiler(Borlan d C++ Builder 6) give
this warning?

It gives the warning because the expression is unreachable. You can
simply ignore the warning or you change the logic. One way would be
change the while loop into a do-while loop.

#include <stdio.h>

int menu( void );

int main( void )
{
int option;

do {
option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
break;

default:
puts("Not a valid option, try again");
}
}while(option != 5);
return 0;
}

int menu( void )
{
int option;

puts("1. List directory");
puts("2. help");
puts("3. time");
puts("4. type menu.c");
puts("5. Quit");
scanf("%d", &option);

return option;
}


Task 2: Create a calculator program.

Below is my humble offering.
=============== =============== =============== =============== =======
#include <stdio.h>

void add(int a, int b);
void subtract(int a, int b);
void multiply(int a, int b);
void divide(int a, int b);

int main( void )
{
int a, b;
char c;

puts("Enter an integer");
scanf("%d", &a);
Add here:
getchar();
puts("Enter an operator");
scanf("%c", &c);
puts("Enter another integer");
scanf("%d", &b);
switch( c ) {
case '*':
multiply(a, b);
break;
case '+':
add(a, b);
break;
case '-':
subtract(a, b);
break;
case '/':
divide(a, b);
break;
default:
puts("Operator is not valid");
}
return 0;
}

void add(int a, int b)
{
printf("Sum is: %d", a+b);
printf("Sum is: %d\n", a+b);
}

void subtract(int a, int b)
{
printf("Answer is: %d", a-b);
printf("Answer is: %d\n", a-b); }

void multiply(int a, int b)
{
printf("Product is: %d", a*b);
printf("Product is: %d\n", a*b); }

void divide(int a, int b)
{
printf("Answer is: %d", a/b);
printf("Answer is: %d\n", a/b); }
=============== =============== =============== ========
Output:

Enter an integer
2
Enter an operator
Enter another integer

Problem: Why does the program not allow the user to input the operator
before
jumping straight to "Enter another integer"? I think I am missing
something very
fundamental here.

When you enter the first number, your input includes the digits for
the number and the newline char, '\n', because you preesed the
enter key. In the first scanf statement, the %d specifier will
stop reading the input buffer when it incounters the '\n' character.
This character remains in the input buffer. In the next scanf
statement you are using the %c specifier. This specifier will not
remove leading whitespace, so the value of the newline char is
read into variable c. The operator,*,/,+ or -, and a newline character
remain in the input buffer. Your next scanf statement, using
the %d, reads the operator. Seeing the operator is not a digit,
it aborts scanning.

The easy solution is to put getchar() after the first scanf
statement which will remove the newline char from the input
buffer and enables you to read the operator into variable c.

--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapi dsys.com (remove the x to send email)
http://www.geocities.com/abowers822/

[Ed Morton]

Buck Rogers wrote:

Hi guys!

Task 1: Write a program which presents a menu with 5 options. The 5th
option quits
the program. Each option should execute a system command using system().

Below is my humble offering.
=============== =============== =============== ======
#include <stdio.h>
#include <stdlib.h>

int menu( void );

int main( void )
{
int option;

while(1) {
option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
exit(0);
default:
puts("Not a valid option, try again");
}
}
return 0;
}
<snip> =============== =============== =============== =============
Problem: My compiler is giving me an "unreachabl e code" warning for line
32(return 0;). The
program runs fine. Why does the compiler(Borlan d C++ Builder 6) give
this warning?

As others have already pointed out, you can't reach that line. Sticking
an "exit" in the middle of your code is pretty nasty in terms of code
structure (it's like having a "goto return;"), and having a loop that
you claim is infinite when it's not, isn't great either. If you have a
loop that termintes under some condition (or the negation of some
condition), then that condition is important enough to be explicitly
named by you so that anyone in future who has to enhance this code
doesn't have to waste time trying to work out the abstraction for that
condition.

To make your code clearer for the next guy who has to work on it (i.e.
more easily maintainable), write it this way:

int main( void )
{
int option;
int prompting = 1;

while(prompting ) {
option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
prompting = 0;
break;
default:
puts("Not a valid option, try again");
}
}
return 0;
}

That will co-incidentally get rid of the compiler warning.

Ed.

[CBFalconer]

Ed Morton wrote:

.... snip ...
To make your code clearer for the next guy who has to work on it
(i.e. more easily maintainable), write it this way:

int main( void )
{
int option;
int prompting = 1;

while(prompting ) {
option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
prompting = 0;
break;
default:
puts("Not a valid option, try again");
}
}
return 0;
}

That will co-incidentally get rid of the compiler warning.

My solution for the same problem:

int main(void)
{
int option;

while(5 != (option = menu())) {
switch(option) {
case 1: system("dir"); break;
case 2: system("help"); break;
case 3: system("time"); break;
case 4: system("type menu.c"); break;
default: puts("Not a valid option, try again"); break;
}
}
puts("Quitting program");
return 0;
}

--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!

[Bruno Desthuilliers]

nrk wrote:

Buck Rogers wrote:

Hi guys!

Task 1: Write a program which presents a menu with 5 options. The 5th
option quits
the program. Each option should execute a system command using system().

Below is my humble offering.
============= =============== =============== ======== (snip code)
Problem: My compiler is giving me an "unreachabl e code" warning for line
32(return 0;). The
program runs fine. Why does the compiler(Borlan d C++ Builder 6) give this
warning?

The only way out of your while ( 1 ) loop is through the exit(0) call, and
that doesn't bring you to the return 0; statement either. Hence the
complaint. Ideally, you should be able to replace the exit(0) call with
return 0;

Or (better IMHO) replace the exit(0) with a break, so 1/code after the
loop may be executed and 2/ the function has a single point of exit
located at the end of the function (which usually makes a more readable
code).

Bruno

[pete]

Buck Rogers wrote:

do {

option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
break;
default:
puts("Not a valid option, try again");
}
}

while (option != 5);

--
pete

[Barry Schwarz]

On Sat, 17 Jan 2004 12:19:41 GMT, nrk
<ra*********@de vnull.verizon.n et> wrote:

Buck Rogers wrote:

Hi guys!

Task 1: Write a program which presents a menu with 5 options. The 5th
option quits
the program. Each option should execute a system command using system().

Below is my humble offering.
=============== =============== =============== ======
#include <stdio.h>
#include <stdlib.h>

int menu( void );

int main( void )
{
int option;

while(1) {
option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
exit(0);
default:
puts("Not a valid option, try again");
}
}
return 0;
}
snip Problem: My compiler is giving me an "unreachabl e code" warning for line
32(return 0;). The
program runs fine. Why does the compiler(Borlan d C++ Builder 6) give this
warning?

The only way out of your while ( 1 ) loop is through the exit(0) call, and
that doesn't bring you to the return 0; statement either. Hence the
complaint. Ideally, you should be able to replace the exit(0) call with

Why? The exit(0) works fine.
return 0; get rid of the existing return statements and the compiler should
be happy.

<<Remove the del for email>>

[nrk]

Barry Schwarz wrote:

On Sat, 17 Jan 2004 12:19:41 GMT, nrk
<ra*********@de vnull.verizon.n et> wrote:

Buck Rogers wrote:

Hi guys!

Task 1: Write a program which presents a menu with 5 options. The 5th
option quits
the program. Each option should execute a system command using system().

Below is my humble offering.
=============== =============== =============== ======
#include <stdio.h>
#include <stdlib.h>

int menu( void );

int main( void )
{
int option;

while(1) {
option = menu();
switch(option) {
case 1:
system("dir");
break;
case 2:
system("help");
break;
case 3:
system("time");
break;
case 4:
system("type menu.c");
break;
case 5:
puts("Quitting program");
exit(0);
default:
puts("Not a valid option, try again");
}
}
return 0;
}
snip Problem: My compiler is giving me an "unreachabl e code" warning for line
32(return 0;). The
program runs fine. Why does the compiler(Borlan d C++ Builder 6) give
this warning?

The only way out of your while ( 1 ) loop is through the exit(0) call, and
that doesn't bring you to the return 0; statement either. Hence the
complaint. Ideally, you should be able to replace the exit(0) call with

Why? The exit(0) works fine.

Except for the complaint from OP's compiler. Simply getting rid of the
return statment may not be an option as the compiler may complain about no
value being returned from main. Of course, you could just happily co-exist
with the compiler diagnostic as well.

-nrk.

return 0; get rid of the existing return statements and the compiler
should be happy.

<<Remove the del for email>>

--
Remove devnull for email