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C++ crash course. Class 7 more operators, expressions, statements. Agenda. More operators! getting values from combining operands Expressions strings of tokens that return a value Statements a line of code in C++. Assignment Operators. Assignment operators

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C++ crash course

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c crash course

C++ crash course

Class 7

more operators, expressions, statements

  • More operators!
    • getting values from combining operands
  • Expressions
    • strings of tokens that return a value
  • Statements
    • a line of code in C++
assignment operators
Assignment Operators
  • Assignment operators
  • An assignment is when we put a value in a variable
  • Left-hand operand must be a non-constlvalue

inti, j, ival;

constint ci = i; // alright; initializing

1024 = ival; // bad

i + j = ival; // bad

ci = ival; // bad

assignment operators1
Assignment Operators
  • Array names are nonmodifiable: you can use subscript and dereference, but you can’t change the array itself

int ia[10];

ia[0] = 0; // OK

*ia = 0; // OK

ia = ???; // bad!

assignment operator
Assignment Operator
  • Result of an assignment is the left-hand operand; type is the type of the left-hand operand

ival = 0; // result: type int value 0

ival = 3.14159; // result: type int value 3

What happens here?

inti = 10, j = 5;

i = j = 4;

assignment operator1
Assignment Operator
  • Result of an assignment is the left-hand operand; type is the type of the left-hand operand

ival = 0; // result: type int value 0

ival = 3.14159; // result: type int value 3

What happens here?

inti = 10, j = 5;

i = j = 4; // assignment is right-associative

assignment operators2
Assignment Operators
  • Assignment is right-associative; other binary operators are left associative

Result of the rightmost assignment is assigned going left

int ival; int *pval;

ival = pval = 0; // error: can’t assign the val of a pointer to an int

string s1, s2;

s1 = s2 = “OK”; // alright; “OK” the char * literal gets converted to a string

assignment operators3
Assignment Operators
  • Assignment has low precedence (happens last)

inti = get_value();

while(i != 42) {

i = get_value();


How can we do this in one line?

assignment operators4
Assignment Operators


while ((i = get_value()) != 42) {

// do something


Without the parentheses, what happens?

while (i = get_value() != 42) {

// do something weird...


assignment operators5
Assignment Operators
  • That means BE CAREFUL whether you are using the comparison operator == or the assignment operator =!

if (i = 42)

if (i == 42)

What’s the difference?

assignment operators6
Assignment Operators

What are i and d after each assignment?

int i; double d;

d = i = 3.5;

i = d = 3.5;

assignment operators7
Assignment Operators
  • What’s the difference here?

if (42 = i) // ...

if (i = 42) // ...

compound assignment operators
Compound Assignment Operators

+= -= *= /= %= // arithmetic operators

(there are also bitwise operators, but we’re not going to talk about those)

a = a op b;

is the same as


But the left-hand operand gets evaluated only once – mostly this doesn’t matter, except when it does (we might see cases of this later)

assignment operators8
Assignment Operators
  • This is illegal. Why?

double dval; intival; int *pi;

dval = ival = pi = 0;

assignment operators9
Assignment Operators
  • These are legal, but they’re not going to do what you might expect them to...
  • What should’ve been written?
  • if (ptr = retrieve_pointer() != 0)
  • if (ival = 1024)
  • ival += ival + 1;
assignment operators10
Assignment Operators
  • Increment and decrement

++, --

inti = 0, j;

j = ++i; // prefix

j = i++; // postfix

Note: prefix operator does less work! (Why?)

assignment operators11
Assignment Operators
  • Combining dereference and increment...

vector<int>::iterator iter = ivec.begin();

while (iter != ivec.end())

cout << *iter++ << endl; // iterator postfix increment

What gets printed out? Why?

assignment operators12
Assignment Operators
  • *iter++ : precedence of postfix increment is higher than the dereference operator

*iter++ is the same as *(iter++)

assignment operators13
Assignment Operators
  • Even though it’s a little confusing, experienced C++ programmers are more likely to use this kind of expression (*iter++)
    • so if you see C++ code, this is probably what you’ll see!
assignment operators14
Assignment Operators
  • What would change if the while loop was written like so?

vector<int>::iterator iter = ivec.begin();

while(iter != ivec.end())

cout << *++iter << endl;

arrow operator
Arrow Operator
  • You’ve seen the dot operator...


// run the same_isbn function of item1


// run the same_animal function of anim1

arrow operator1
Arrow Operator
  • What if instead of having an explicit Animal_type, we instead have a pointer?

Animal_type *ap = &anim1;


But you have to be careful – what does this do?


arrow operator2
Arrow Operator
  • Thus we have the arrow operator ->


It’s just syntactic sugar, but makes it much easier not to make mistakes!

arrow operator3
Arrow Operator
  • Let’s write code:

Define a vector of pointers to strings

Read the vector, printing each string and its corresponding size

arrow operator4
Arrow Operator
  • Assume iter is a vector<string>::iterator
  • Which of the following are legal?
  • What do they do?
  • *iter++;
  • *iter.empty();
  • ++*iter;
  • (*iter)++;
  • iter->empty();
  • iter++->empty();
conditional operator
Conditional Operator
  • Ternary operator – takes 3 arguments

cond ? expr1 : expr2;

equivalent to

if(cond) {


} else {



conditional operator1
Conditional Operator

int i = 10, j = 20, k = 30;

int maxVal = i > j ? i : j;

What’s maxVal after this?

conditional operator2
Conditional Operator
  • It’s a useful convention, but don’t abuse it

int max = i > j

? i > k ? i : k

: j > k ? j : k;

What does the above do?

conditional operator3
Conditional Operator
  • Equivalent to this...

int max = i;

if (j > max)

max = j;

if (k > max)

max = k;

conditional operator4
Conditional Operator
  • Fairly low precedence
  • In an output expression:

cout << (i < j ? i : j); // works as expected

cout << (i < j) ? i : j; // prints 1 or 0 – why?

cout << i < j ? i : j; // error – why?

conditional operator5
Conditional Operator
  • Writing code:

Prompt the user for a pair of numbers, say which is smaller

conditional operator6
Conditional Operator
  • Write code

Process elements in a vector<int>, replacing any value that’s odd with twice its value

sizeof operator
sizeof Operator
  • Returns a value of type size_t
  • Size in bytes of an object or type name – how much space it’s taking up in memory

sizeof (type name);

sizeof (expr);

sizeof expr;

(Remember when I said that doing ptr++ will cause it to move forward by a certain amount in memory?)

sizeof operator1
sizeof Operator
  • Can be used in multiple ways...

Animal_typeanim, *pa;



sizeof *pa;

sizeof operator2
sizeof Operator
  • Evaluating sizeofexprdoes not evaluate the expression

sizeof *p; // will work even if p is a null pointer or invalid address!

sizeof operator3
sizeof Operator
  • Result of sizeof depends on the type involved
    • sizeof char or an expression of type char is 1
    • sizeof a reference type returns the size of the memory necessary to contain an object of the referenced type
    • sizeof a pointer returns the size needed to hold a pointer; to obtain the size of the object, the pointer must be dereferenced
    • sizeof an array is equivalent to taking sizeof the element type, times the number of elements in the array
sizeof operator4
sizeof Operator
  • We can figure out the number of elements in an array this way!
    • (since there’s no array.size())


sizeof operator5
sizeof Operator
  • What’s the output of this program going to be?

[[ size of an int: 4 bytes; size of an int *: 8 bytes ]]

int x[10]; int *p = x;

cout << sizeof(x) / sizeof(*x) << endl;

cout << sizeof(p) / sizeof(*p) << endl;

compound expressions
Compound Expressions
  • An expression with two or more operators is a compound expression
  • Must understand how precedence and associativity work in order to evaluate expressions
  • Precedence doesn’t necessarily specify order of evaluation, but it does determine what the answer will be
  • We saw this yesterday:
    • How does this get evaluated?

6 + 3 * 4 / 2 + 2;

Parentheses can override precedence

((6 + ((3 * 4) / 2) + 2)


((6 + 3) * 4) / 2 + 2

  • Specifies how to group operators at the same precedence level

Assignment is right associative

Arithmetic is left associative

  • Parenthesize these:

ival = jval = kval = lval;

ival * jval / kval * lval;

  • When you know what the precedence is, it’s okay not to parenthesize – when you’re not sure, parenthesizing can help make sure that the program does what you want it to
  • Helpful in debugging: getting a weird result? Add parentheses around the things you thought were getting evaluated first
  • You can also look up the precedence for all the operators, but unless you’re an expert C++ programmer, it’s unlikely that you’ll be able to remember all of them
  • Useful to know for any programming language
new and delete
new and delete
  • Has to do with allocating memory
  • We talked about this with dynamic arrays
  • We can also use it for single objects

inti; // named, unitializedint

int *pi = new int; // pi points to a dynamically allocated, unnamed, uninitialized int

new and delete1
new and delete
  • We need to initialize dynamically allocated objects too, even if they’re not named

int i(1024);

int *pi = new int(1024);

string s(10, ‘9’);

string *ps = new string(10, ‘9’);

Must use direct-initialization syntax, rather than copy-initialization, to initialize dynamically allocated objects

new and delete2
new and delete
  • Without an explicit initializer, a dynamically allocated object is initialized in the same way as a variable inside a function
  • If there’s a default constructor, uses that; if it’s built-in, it’s uninitialized

After each line, which is / are initialized?

string *ps = new string;

int *pi = new int;

new and delete3
new and delete
  • We can also value-initialize a dynamically allocated object
  • Default constructor, or basic type to 0

string *ps = new string();

int *pi = new int();

cls *pc = new cls();

new and delete4
new and delete
  • So that’s new...
  • ...why do we have delete?
  • Back in the day using new too many times could easily exhaust all your memory, causing a bad_allocexception
  • In order not to have our memory used up, we should free it using the delete expression

delete pi;

This will ONLY WORK for dynamically allocated variables; for all others it’s illegal – the behavior is undefined

new and delete5
new and delete

int i;

int *pi = &i;

string str = “dwarves”;

double *pd = new double(33);

Which of these is safe?

delete str;

delete pi;

delete pd;

new and delete6
new and delete

int i;

int *pi = &i;

string str = “dwarves”;

double *pd = new double(33);

Which of these is safe?

delete str;

delete pi; // Watch out: some compilers will accept this even though it’s illegal!

delete pd;

new and delete7
new and delete
  • Once you delete a pointer to an object, you’ve deleted the object, but the pointer still hangs around, pointing to that spot in memory
  • Called a dangling pointer
  • Watch out for these!
    • Best to set any pointers you’ve deleted to 0 (null pointer)
new and delete8
new and delete

Which of these are okay? Which are illegal / error-producing?

  • vector<string> svec(10);
  • vector<string> *pvec1 = new vector<string>(10);
  • vector<string> **pvec2 = new vector<string>[10];
  • vector<string> *pv1 = &svec;
  • vector<string> *pv2 = pvec1;
  • delete svec;
  • delete pvec1;
  • delete [] pvec2;
  • delete pv1;
  • delete pv2;
type conversions
Type Conversions
  • We discussed these a little bit yesterday: typecasting
  • Types of operands determine whether an expression is legal
    • if the types are related in C++, then there’s a conversion

int ival = 0;

ival = 3.541 + 3; // compiles, sometimes gives warning

type conversions1
Type Conversions
  • C++ has a set of conversions that make the operands the same before doing the arithmetic
  • These conversions are done automatically by the compiler, called implicit type conversions
  • Built-in conversions preserve precision if possible
type conversions2
Type Conversions
  • When is an implicit type conversion going to occur?
    • Expressions with mixed-type operands:


double dval;

ival >= dval; // ival converted to double

    • Expression used as a condition goes to bool


if (ival)

while (cin)

    • An expression used to initialize or assign to a variable is converted to the type of the variable

intival = 3.14;

int *ip;

ip = 0;

type conversions3
Type Conversions
  • Simplest conversion: integral promotions

char, signed char, unsigned char, short, and unsigned short can all get promoted to int

If the values don’t fit, promoted to unsigned int

For bools:

false -> 0

true -> 1

type conversions4
Type Conversions
  • Pointer conversions

When we use an array, the array is converted to a pointer to the first element:


int * ip = ia;

type conversions5
Type Conversions
  • Enumeration conversions:

// point2d is 2, point2w is 3, point3d is 3, point3w is 4

enum Points { point2d = 2, point2w,

point3d = 3, point3w };

constsize_tarray_size = 1024;

intchunk_size = array_size * point2w;

int array_3d = array_size * point3d;

type conversions6
Type Conversions
  • What conversions are happening here?

float fval;

double dval;


  • if (fval)
  • dval = fval + ival;
  • dval + ival + cval;
type conversions7
Type Conversions
  • We can also explicitly cast one type to another
  • This is necessary sometimes, but dangerous
    • part of memory is suddenly being re-interpreted!
  • Really useful in terms of arithmetic, which is pretty much the only time you’ll need it

How can we make integer division work?