CS100 Lecture 3

Operators and Control Flow I

Contents

• Operators
  • +, -, *, /, %
  • Compound assignment operators
  • Signed integer overflow
  • ++ and --
• Control flow
  • if-else
  • while
  • for

Operators

The calculator

Accept input of the form x op y, where x and y are floating-point numbers and op '+', '-', '*', '/' . Print the result.

#include <stdio.h>

int main(void) {
  double x, y;
  char op;
  scanf("%lf %c %lf", &x, &op, &y);
  if (op == '+')
    printf("%lf\n", x + y);
  else if (op == '-')
    printf("%lf\n", x - y);

  else if (op == '*')
    printf("%lf\n", x * y);
  else if (op == '/')
    printf("%lf\n", x / y);
  else
    printf("Invalid operator.\n");
  return 0;
}

Overview of arithmetic operators

+, -, *, /, %

• + and - have two versions: unary (+a, -a) and binary (a+b, a-b).

  • The unary +/- and binary +/- are different operators, although they use the same notation.

• Operator precedence:

  unary +, unary - *, /, % binary +, binary -

  e.g. a + b * c is interpreted as a + (b * c), instead of (a + b) * c.

  We will talk more about operator precedence later.

Binary +, - and *, /

a + b, a - b, a * b, a / b

Before the evaluation of such an expression, the operands (a, b) undergo a sequence of type conversions.

• The detailed rules of the conversions are very complex,
  • including promotions, conversions between signed and unsigned types, conversions between integers and floating-point types, etc.
  • We only need to remember some common ones.
• In the end, the operands will be converted to a same type, denoted T. The result type is also T.

Binary +, - and *, /

a + b, a - b, a * b, a / b

If any one operand is of floating-point type and the other is an integer, the integer will be implicitly converted to that floating-point type.

Example:

double pi = 3.14;
int diameter = 20;
WhatType c = pi * diameter; // What is the type of this result?

Binary +, - and *, /

a + b, a - b, a * b, a / b

If any one operand is of floating-point type and the other is an integer, the integer will be implicitly converted to that floating-point type.

Example:

double pi = 3.14;
int diameter = 20;
double c = pi * diameter; // 62.8

The value of diameter is implicitly converted to a value of type double. Then, a floating-point multiplication is performed, yielding a result of type double.

* Does this rule make sense?

Binary +, - and *, /

a + b, a - b, a * b, a / b

If any one operand is of floating-point type and the other is an integer, the integer will be implicitly converted to that floating-point type.

Example:

double pi = 3.14;
int diameter = 20;
double c = pi * diameter; // 62.8

The value of diameter is implicitly converted to a value of type double. Then, a floating-point multiplication is performed, yielding a result of type double.

* Does this rule make sense? - Yes, because .

Binary +, - and *, /

a + b, a - b, a * b, a / b

If any one operand is of floating-point type and the other is an integer, the integer will be implicitly converted to that floating-point type, and the result type is that floating-point type.

Similarly, if the operands are of types int and long long, the int value will be implicitly converted to long long, and the result type is long long.

Division: a / b

Assume a and b are of the same type T (after conversions as mentioned above).

• Then, the result type is also T.

Two cases:

• If T is a floating-point type, this is a floating-point division.
• If T is an integer type, this is an integer division.

Division: a / b

Two cases:

• If T is a floating-point type, this is a floating-point division.
  • The result is no surprising.
• If T is an integer type, this is an integer division.
  • The result is truncated towards zero (since C99 and C++11) .
  • What is the result of 3 / -2?

Let a and b be two integers.

• What is the difference between a / 2 and a / 2.0?
• What does (a + 0.0) / b mean? What about 1.0 * a / b?

Division: a / b

If T is an integer type, this is an integer division.

• The result is truncated towards zero (since C99 and C++11) .
• What is the result of 3 / -2?
  • -1.5 truncated towards zero, which is -1.

What is the difference between a / 2 and a / 2.0?

• a / 2 yields an integer, while a / 2.0 yields a double.

What does (a + 0.0) / b mean? What about 1.0 * a / b?

• Both use floating-point division to compute . The floating-point numbers 0.0 and 1.0 here cause the conversion of the other operands.

Remainder: a % b

Example: 15 % 4 == 3.

a and b must have integer types.

If a is negative, is the result negative? What if b is negative? What if both are negative?

Remainder: a % b

Example: 15 % 4 == 3.

a and b must have integer types.

If a is negative, is the result negative? What if b is negative? What if both are negative?

For any integers a and b, the following always holds:

(a / b) * b + (a % b) == a

Compound assignment operators

+=, -=, *=, /=, %=

• a op= b is equivalent to a = a op b.
• e.g. x *= 2 is equivalent to x = x * 2.
• [Best practice] Learn to use these operators, to make your code clear and simple.

Signed integer overflow

If a signed integer type holds a value that is not in the valid range, overflow is caused.

Suppose int is 32-bit and long long is 64-bit.

Do the following computations cause overflow?

int ival = 100000; long long llval = ival;
int result1 = ival * ival;
long long result2 = ival * ival;
long long result3 = llval * ival;
long long result4 = llval * ival * ival;

Signed integer overflow

Suppose int is 32-bit and long long is 64-bit.

Do the following computations cause overflow?

int ival = 100000; long long llval = ival;
int result1 = ival * ival;               // (1) overflow
long long result2 = ival * ival;         // (2) overflow
long long result3 = llval * ival;        // (3) not overflow
long long result4 = llval * ival * ival; // (4) not overflow

(1) .

(2) The result type of the multiplication ival * ival is int, which causes overflow. This is not affected by the type of result2.

Signed integer overflow

Suppose int is 32-bit and long long is 64-bit.

Do the following computations cause overflow?

int ival = 100000; long long llval = ival;
int result1 = ival * ival;               // (1) overflow
long long result2 = ival * ival;         // (2) overflow
long long result3 = llval * ival;        // (3) not overflow
long long result4 = llval * ival * ival; // (4) not overflow

(3) Since llval is of type long long, the value of ival will be implicitly converted to long long, and then the multiplication yields a long long value.

(4) * is left-associative, so the expression a * b * c is interpreted as (a * b) * c.

We will talk about associativity in later lectures.

Undefined behavior

Signed integer overflow is undefined behavior: There are no restrictions on the behavior of the program. Compilers are not required to diagnose undefined behavior (although many simple situations are diagnosed), and the compiled program is not required to do anything meaningful.

• It may yield some garbage values, or zero, or anything else;
• or, this statement may be removed if the compiler is clever enough;
• or, the program may crash;
• or, any other results beyond imagination.

More on undefined behaviors in recitations.

Unsigned integers never overflow

Unsigned integer arithmetic is always performed modulo , where is the number of bits in that integer type.

For example, for unsigned int (assuming it is 32-bit)

• adding one to gives because , and
• subtracting one from gives because .

* "wrap-around"

Increment/decrement operators

Unary operators that increment/decrement the value of a variable by 1.

Postfix form: a++, a--

Prefix form: ++a, --a

• a++ and ++a increment the value of a by 1.
• a-- and --a decrement the value of a by 1.

Increment/decrement operators

Unary operators that increment/decrement the value of a variable by 1.

Postfix form: a++, a--

The result of the postfix increment/decrement operators is the value of a before incrementation/decrementation.

* What does "result" mean?

Increment/decrement operators

Unary operators that increment/decrement the value of a variable by 1.

Postfix form: a++, a--

The result of the postfix increment/decrement operators is the value of a before incrementation/decrementation.

int x = 42;
printf("%d\n", x++); // x becomes 43, but 42 is printed.
int y = x++; // y is initialized with 43. x becomes 44.

Increment/decrement operators

Unary operators that increment/decrement the value of a variable by 1.

Prefix form: ++a, --a

The result of the prefix increment/decrement operators is the value of a after incrementation/decrementation.

int x = 42;
printf("%d\n", ++x); // x becomes 43, and 43 is printed.
int y = ++x; // y is initialized with 44. x becomes 44.

Control flow

if-else

Conditionally executes some code.

Syntax:

• (1) if (condition) statementT
• (2) if (condition) statementT else statementF

where condition is an expression of arithmetic type or pointer type.

We will talk about pointers in later lectures.

If condition compares not equal to the integer zero, statementT is executed.

In (2), if condition compares equal to the integer zero, statementF is executed.

if-else

• (1) if (condition) statementT
• (2) if (condition) statementT else statementF

statementT, statementF: Either a statement or a block (any number of statements surrounded by {}).

• If more than one statements are to be executed, they must be surrounded with a pair of braces {}.

int i = 42;
if (i == 42)
  printf("hello");
  printf("world\n"); // This is NOT a part of statementT!
                     // This code is incorrectly indented.

if-else

• (1) if (condition) statementT
• (2) if (condition) statementT else statementF

statementT, statementF: Either a statement or a block (any number of statements surrounded by {}).

• If more than one statements are to be executed, they must be surrounded with a pair of braces {}.

int i = 42;
if (i == 42) {
  printf("hello");
  printf("world\n");
}

The calculator example

double x, y;
char op;
scanf("%lf %c %lf", &x, &op, &y);
if (op == '+') {
  printf("%lf + %lf == %lf\n", x, y, x + y);
} else if (op == '-') {
  printf("%lf - %lf == %lf\n", x, y, x - y);
} else if (op == '*') {
  printf("%lf * %lf == %lf\n", x, y, x * y);
} else if (op == '/') {
  printf("%lf / %lf == %lf\n", x, y, x / y);
} else {
  printf("Unknown operator!\n");
}

The calculator example

If there is only one statement, the braces can be omitted.

double x, y;
char op;
scanf("%lf %c %lf", &x, &op, &y);
if (op == '+')
  printf("%lf + %lf == %lf\n", x, y, x + y);
else if (op == '-')
  printf("%lf - %lf == %lf\n", x, y, x - y);
else if (op == '*')
  printf("%lf * %lf == %lf\n", x, y, x * y);
else if (op == '/')
  printf("%lf / %lf == %lf\n", x, y, x / y);
else
  printf("Unknown operator!\n");

The calculator example

Nested if statements are possible:

double x, y;
char op;
scanf("%lf %c %lf", &x, &op, &y);
if (op == '+')
  printf("%lf + %lf == %lf\n", x, y, x + y);
else if (op == '-')
  printf("%lf - %lf == %lf\n", x, y, x - y);
else if (op == '*')
  printf("%lf * %lf == %lf\n", x, y, x * y);
else if (op == '/') {
  if (y == 0)
    printf("Error: division by zero.\n");
  else
    printf("%lf / %lf == %lf\n", x, y, x / y);
} else
  printf("Unknown operator!\n");

Dangling else

The else is always associated with the closest preceding if.

if (i == 1)
  if (j == 2)
    printf("aaa\n");
else // Incorrectly indented! This `else` is associated with `if (j == 2)`.
  printf("bbb\n");

If you want this else to be associated with if (i == 1), add a pair of braces.

if (i == 1) {
  if (j == 2)
    printf("aaa\n");
} else
  printf("bbb\n");

The formatter tells you the truth.

Unlike Python, indentations are not part of the C syntax. The following code is completely legal, but with extremely low readability.

int a,b;scanf("%d%d",&a,&b);int s=a+b;printf("%d\n",
s);if(a==1)if(b==2)printf("aaa\n");else printf("b\n");

[Best practice] Use a formatter! If you use VSCode, press Shift+Alt+F (Windows), ⇧⌥F (MacOS) or Ctrl+Shift+I (Linux) to format the code.

• Other editors may also have this functionality.

The formats are customizable. Explore it on your own if you are interested.

while

Executes some code repeatedly under certain condition.

Syntax:

while (condition) loop_body

where condition is an expression of arithmetic type or pointer type, and loop_body is a statement or a block.

while

Read an integer n, followed by n integers. Print the sum of the n integers.

int n;
scanf("%d", &n);
int sum = 0;
while (n > 0) {
  int x;
  scanf("%d", &x);
  sum += x;
  --n;
}
printf("%d\n", sum);

while

while (n > 0) {
  // loop body
  --n;
}

How many times is this loop body executed?

After the loop, what is the value of n?

while

while (n > 0) {
  // loop body
  --n;
}

How many times is this loop body executed?

• n.

After the loop, what is the value of n?

• 0, which makes n > 0 not satisfied.

while

A simpler and common way to write this loop:

while (n--) {
  // loop body
}

How many times is this loop body executed?

After the loop, what is the value of n?

while

A simpler and common way to write this loop:

while (n--) {
  // loop body
}

How many times is this loop body executed?

• n.

After the loop, what is the value of n?

• -1. When n == 0, the result of n-- is 0, which not only makes the loop terminated but also decrements n by 1.

break

break; causes the enclosing loop to terminate.

while (n--) {
  int x;
  scanf("%d", &x);
  // If x == 42, the loop is terminated and control goes to (*).
  if (x == 42)
    break;
  sum += x;
}
// (*)
printf("%d\n", sum);

continue

continue; causes the remaining portion of the enclosing loop body to be skipped (in the current iteration).

while (n--) {
  int x;
  scanf("%d", &x);
  // If x == 42, the rest of the loop body is skipped and control goes to (*).
  if (x == 42)
    continue;
  sum += x;
  // (*)
}
printf("%d\n", sum);

break vs continue

while (n--) {
  int x;
  scanf("%d", &x);
  if (x == 42)
    break; // goto (*)
  sum += x;
}
// (*)
printf("%d\n", sum);

while (n--) {
  int x;
  scanf("%d", &x);
  if (x == 42)
    continue; // goto (*)
  sum += x;
  // (*)
}
printf("%d\n", sum);

for

Syntax: for (init_clause; condition; iteration_expr) loop_body

• init_clause may be an expression or a declaration (since C99).
• condition is an expression of arithmetic type or pointer type. It is evaluated before the loop body.
• iteration_expr is an expression evaluated after the loop body.
• loop_body is either a statement or a block.

for

Typical usage: execute the body for n times.

The following are equivalent.

while (n--) {
  int x;
  scanf("%d", &x);
  sum += x;
}

for (int i = 0; i < n; ++i) {
  int x;
  scanf("%d", &x);
  sum += x;
}

for

Syntax: for (init_clause; condition; iteration_expr) loop_body

It is equivalent to

{
  init_clause;
  while (condition) {
    loop_body
    iteration_expr;
  }
}

for

A loop with iteration variable i incremented by 2 every time:

for (int i = 0; i < n; i += 2)
  // ...

Loop until :

for (int i = 0; i * i <= n; ++i) // We don't need sqrt.
  // ...

A loop with iteration variable i going downward from n to 0:

for (int i = n; i >= 0; --i)
  // ...

* Can we use unsigned here?

for

What will happen?

for (unsigned i = n; i >= 0; --i)
  // ...

for

What will happen?

for (unsigned i = n; i >= 0; --i)
  // ...

i is an unsigned integer, so i >= 0 holds forever.

• When i == 0, --i makes the value of i become 4294967295 (if unsigned is 32-bit).
• An unsigned integer never overflows. Unsigned integer arithmetic is performed modulo .

for

[Best practice] Always declare and initialize the iteration variable inside the for statement, if possible.

main() {
  int i, j;
  /* ... */
  for (i = 0; i < n; ++i)
    for (j = 0; j < m; ++j)
      /* ... */
}

int main(void) {
  // ...
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      // ...
}

Summary

Operators

• a + b, a - b, a * b, a / b: a and b are converted to a common type T according to some rules, and the result type is also T.
• Integer division: The result is truncated towards zero.
• Remainder: (a / b) * b + (a % b) == a always holds.
• +=, -=, *=, /=, %=: a op= b a = a op b
• Signed integer overflow: undefined behavior.
• Unsigned integer arithmetic: modulo , never overflows.
• a++ and a-- return the original value. ++a and --a return the new value.

Summary

Control flow

• if-else
• while, break, continue
• for

Exercises

1. Write a program that reads an unknown number of integers from input until 0 is entered. Print the sum of these integers.
2. Make the calculator a loop that runs repeatedly, instead of handling only one input.

Notes

Similar conversions also happen for operands with types int + long, long + long long, unsigned + unsigned long long, etc. However, if any operand is of type with rank less than rank of int, integer promotion is first applied. For example, if a is of type signed char and b is of type short, both of them are converted to int first and the result type is int, not short.

Integer promotion is also applied to the operand of unary + and -. For example, +x is of type int if x is a short variable.

Before C99 and C++11, the result of integer division is rounded in an implementation-defined direction.