Help on getting started with C++

Black_Rose

· c = b/a
· a = b
· a = b = 5
· c = ++a + b--

a, b & c are integers..

a = 8, b = 9;

c = b/a;

c = 9 / 8 = 1.125

Since c is an integer, its floating point precision will be deducted, means int datatype doesn't take the digits after the decimal point.

If you want the floating point precision use another datatype for c like float or double.

& the flow goes as following:

a = b; // the value of b replaces the value of a, now a = 9;

a = b = 5; // now 5 replaces the values of a & b now both equal 5

c = ++a + b--;

++ means incrementing the value of the variable by 1
-- means decrementing the value of the variable by 1

but their place in the next to the value differ the time the operation occur, if the they're on the left side the operation occur then do the instruction (line).

means a & b now equal 5, ++a = 6, b-- = 4, but b won't decrement till the line is done.

So, c = (++a) 6 + (b--) 5;
c = 11;
b = 4; //after the operation is done.

Black_Rose

#include stdio.h

void main (void)

int a, b, c;

a = 8;
b = 9;
c = b/a;

printf (c=b/a);
Whilst on the topic, what does the printf command do, the scanf command?

this suppose to show the value of c.

but the printf() function isn't declared right.

printf("%d",c) OR printf("%d",b/a)

printf() shows the value of strings, now you want to show an integer so we use "%d" to show the value of c & you can't assign a value "c=b/a" in printf().

scanf() takes input from user, like:
int n;
scanf("%d",&n);
this waits for user input & press "Enter" button then put the value in "&n"

&n : is the address of variable n in the memory, the bits value of assigned to its address, that's why we dont just put scanf("%d",n).

& the "%d", means read it as an integer.


I hope this is helpful for your query, feel free to ask and you can know more #-Link-Snipped-# & there are plenty of tutorials on a web search for "C++ tutorials"