C Program Learning Part-5
Decision Making and Branching
language possesses the following statements:
* statement
*switch statement
*Conditional statement
*goto statement
Decision Making With IF Statement:
It takes the following
form:
if(test expression)
SIMPLE IF STATEMENT
The general form of a simple if statement is:
if(test expression)
{
statement block;
}
statement-x;
main()
{
int a, b, c, d;
float ratio;
printf(“Enter four integer
number.\n”);
scanf(“%d %d %d %d”, &a,
&b, &c, &d);
if((c-d)!=0)
{
ratio=
(float)(a+b)/ (float)(c-d);
printf(“Ratio =
%f”, ratio);
}
}
A Program by using previous Format:
main()
{
int count, i;
float weight, height;
count = 0;
printf(“Enter weight and height
for 10 boys. \n”);
for(i = 1; i <= 10; i++)
{
scanf(“%f %f”, &weight,
&height);
if( weight < 50 &&
height > 170)
{
count = count + 1;
}
}
printf(“Number of boys with
weight < 50 kg \n”);
printf(“and height > 170 cm =
%d”,count);
}
THE IF………….. ELSE STATEMENT
The general form is
if(test expression)
{
true block statement(s):
}
else
{
false block statement(s):
}
statement-x
if((c-d)!=0)
{
ratio= (float)(a+b)/ (float)(c-d);
printf(“Ratio = %f”, ratio);
}
else
printf(“c-d is zero”);
NESTING OF
IF………..ELSE STATEMENTS
if(test condition -1)
{
if(test condition -2)
{
statement -1;
{
else
}
statement -2;
}
}
else
{
statement -3;
}
statements -x;
On Application Format:
if(sex is female)
{
if(balance > 5000)
{
bonus = 0.05 * balance;
}
else
{
bonus = 0.02 * balance;
}
}
else
{
bonus = 0.02 * balance;
}
balance = balance + bonus;
Format:
main()
{
float a, b, c;
printf(“Enter three values: \n”);
scanf(“%f %f %f”, &a, &b,
&c);
printf(“The largest value is: ”);
if(a > b)
{
if(a > c)
{
printf(“%f”,a);
}
else
printf(“%f”,c);
}
else
{
if(c > b)
printf(“%d”,c);
else printf(“%d”,b);
}
}
Dangling Else Problem
This occurs when a matching else is not available for an if.
To avoid this always match an else to the most recent unmatched if in the
current block.
Have logic error:
x= 10;
y= 2;
if( x > 5)
if( y > 5)
printf(“ x and y are > 5”);
else
printf(“x is <= 5.”);
Correctness:
x= 10;
y= 2;
if( x > 5)
{
if( y > 5)
printf(“ x and y are > 5”);
}
else
printf(“x is <= 5.”);
|
Consumption Units |
Rate of Charge |
|
0-200 |
Tk 0.05 per unit |
|
201-400 |
Tk 100 per unit |
|
401-600 |
Tk 230 per unit |
|
601 and above |
Tk 390 per unit |
Now write a program that reads the customer number and power
consumed and prints the amount to be paid by the customer.
Solution:
main()
{
int units, customer;
float charges;
printf(“Enter Customer NO. and
UNITS Consumed\n”);
scanf(“%d %d”, &customer,
&units);
if(units<=200)
charges = 0.5 * units; else
if(units <= 400)
{
charges = 100 + 0.65 * (units -
200);
}
else if(units <= 600)
{
charges = 230 + 0.8 * (units -
400);
}
else
{
charges = 390 + 1.0 * (units -
400);
}
printf(“Customer number = %d and
Charges = %.2f”,customer, units);
}
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