CS2100: Arrays and TypeStruct

Arrays

It's illegal to assign an array to another array.

#define N 10

int source[N] = { 10, 20, 30, 40, 50 };

int dest[N];

dest = source;  // illegal!

This is due to the array name being a constant pointer, it points to the first element of the array

This cannot be changed.

If we want to do this, we have to write a loop...

All array names point to the first element of the array memory.

We don't need to pass the pointer of the array to a function since the name itself is already an array.

Strings

All strings are character array that ends with "\0" which is a null character with a ASCII value of zero.

Without this character, it's just an array of character.

1. char fruit_name[] = "apple";

2. char fruit_name[] = {'a','p','p','l','e','\0'};

Reading Strings from stdin:

fgets(str, size, stdin) // reads size – 1 char,

                 // or until newline

Note: It will read a newline character, thus we need to replace it will "\0" 

e.g the string "eat" will be {'e','a','t','\n','\0'}

scanf("%s", str);  // reads until white space

Printing:

puts(str);  // terminates with newline

printf("%s\n", str);

Structures

A type is not a variable and needs to be defined before declaring variables for that type.

No memory is allocated to a type.

Declaration:

Initialisation:

result_t result1 = { 123321, 93.5, 'A' };

We can also use the dot operator, it acts like an object in java

result_t result2;

result2.stuNum = 456654;

result2.score = 62.0;

result2.grade = 'D';

However, when reading, we need to use the address when reading the values.

scanf("%d %f %c", &result1.stuNum, &result1.score,
                  &result1.grade);

Use the name to refer to the entire structure and the dot operator to refer to an individual member.

Unlike arrays, we can do assignments. ie result2 = result1;

Structures can also be returned as return types from functions.

The arrow operator:
Instead of using the dot, we can use "->" as a shortcut

(*player_ptr).name is equal to player_ptr->name

(*player_ptr).age is equal to player_ptr->age 

We cannot forget the brackets as the dot is higher precedence than * thus removing the bracket will cause the dot to be evaluated first. Ie. it be read as *(player_ptr.name) instead

Note: When passing to functions,
if we want to edit the data, we need to pass in the memory referenced by &(Name of data)
and in the function, referenced by *(Name of data) and assess it by (*nameofdata).variable1.
However, if we are not altering the data, there's no need to pass in the reference.


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