From ab9b467243c6eb1860e63cb7b7fa85885447d803 Mon Sep 17 00:00:00 2001
From: loshprung <lshprung@scu.edu>
Date: Mon, 13 Jan 2020 11:32:51 -0800
Subject: post-class 01-13-20

---
 01-13.md | 138 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-
 1 file changed, 137 insertions(+), 1 deletion(-)

(limited to '01-13.md')

diff --git a/01-13.md b/01-13.md
index 20e834f..2441c7f 100644
--- a/01-13.md
+++ b/01-13.md
@@ -1 +1,137 @@
-# Notes 01/13/2020
+Notes 01/13/2020
+
+> Review from COEN10
+# Pointers
+
+	int x;
+	x = 6;
+	int *px;
+	px = &x;
+
+- The pointer `*px` is "pointing" to the memory address where x is located (`&x`)
+- The pointer must be assigned a data type (in this case `int`) that matches the memory type that it is pointing to
+
+
+	x = 5;
+	*px = 1
+	printf("%d", x);
+
+
+- The code above will print `1` because `*px` (the value in the memory address that px is pointing to) was changed to 1, so x was changed to 1
+- You can change where the pointer is pointing to: `px++` changes the pointer to point to the next memory address
+	- This is useful when dealing with arrays:
+
+
+	int x[5] = {1, 2, 3, 4, 5}
+	int *px;
+	px = &x[0]; //can also be written px = x
+	printf("%d", *px);
+	px++;
+	printf("%d", *px);
+
+
+The code above would print `1` and then `px++` would move the pointer to point to `x[1]` and the second number printed would be `2`
+
+
+	int i;
+	px = &x[0];
+	for(i = 0; i < 5; i++){
+		printf("%d\n", *px);
+		px++;
+	}
+
+
+The code above would result in:
+
+
+	1
+	2
+	3
+	4
+	5
+
+
+Another Example:
+
+
+	int x = 2, y = 2;
+	int *p, *q;
+	p = &y;
+	q = &y;
+	//(p == q) is false (comparison of addresses)
+	//(*p == *q) is true (comparison of values)
+
+
+Visual Representation of Memory:
+
+	int x[7] = {1, 2, 3, 4, 5, 6, 7};
+	int *px;
+	px = &x[0];
+
+`x[]` in memory
+
+| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 
+| - | - | - | - | - | - | - | 
+| px |  |  |  |  |  |  | 
+
+	px+=3;
+
+| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 
+| - | - | - | - | - | - | - | 
+|  |  |  | px |  |  |  | 
+
+---
+
+# Using Pointers with Strings
+
+Example:
+
+	char my_string[] = "String!"
+	char *p = my_string //Single line pointer definition. can also be written *p = &my_string[0]
+
+Visual Representation of `my_string[]` in memory:
+
+| S | T | R | I | N | G | ! | \0 |
+| - | - | - | - | - | - | - | -  |
+| p |   |   |   |   |   |   |    |
+
+We can print the string like this:
+
+	//don't need i
+	while(*p != '\0'){
+		printf("%c", *p);
+		p++;
+	}
+
+The above code would print `STRING!`
+
+---
+
+# Pointers and Functions
+
+- A function can also receive an array as an argument in the form of a pointer
+	- For Example:
+
+
+	void my_function(char *c){
+		...
+	}
+
+
+- The function would receive a memory address pointing to a char. It would be beneficial to pass a memory address pointing to a string. 
+	- Let's flesh-out the example:
+
+
+	void my_function(char *c){
+		printf("%s\n", *c);
+		c++;
+		printf("%s\n", *c);
+	}
+
+
+- Let's say we pass a memory address `&my_string[0]` and my_string is defined as `char my_string[] = "Hello World"`
+	- The code would output:
+
+
+	Hello World
+	ello World
-- 
cgit