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16 changes: 16 additions & 0 deletions assignment7.txt
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1. For an array {1, 2, 3, 4}
a. ++*p will get the variable located at *p then increment it. if *p is at index 0, new array would be {2, 2, 3, 4} with *p at index 0
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@oldclesleycode oldclesleycode Jul 11, 2016

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The expression ++*p has two operators of same precedence, so compiler looks for associativity. Associativity of operators is right to left. Therefore the expression is treated as ++(*p). The expression *p++ is treated as *(p++) as the precedence of postfix ++ is higher than *. The expression *++p has two operators of same precedence, so compiler looks for associativity. Associativity of operators is right to left. Therefore the expression is treated as *(++p).

b. *p++ will increment p (move the index forward once), then get the value of it. If *p is at index 0, then the array will be {1,2,3,4} with *p at index 1
c. *++p is equivalent to *p++ as the prefix ++ and * evaluate right to left.
2. No, it depends on the type of operator.
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right!

3. Pointers are used for functions that need to actually change their arguments or for returning an array of values
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-Pointers are more efficient in handling arrays and data tables
-They can be used to return multiple values from a function via function arguments
-Pointers permit references to functions
-Pointers allow C to support dynamic memory management
-Pointers reduce length and complexity of programs

4.1 char[]
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Close! We're looking for the data type, so it's char*

4.2 invalid
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This is actually valid! "xyz" is just an array of characters, so the "xyz"[1] is just accessing "y". Then you just subtract 'y' from it to get 0. This is because, if you can recall, a char is just a value mapped to a character!

4.3 int 0
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'\0' is just a NULL terminator and by definition, NULL is equal to 0. So this would evaluate as true, which is 1 in C.

4.4 int 10
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right!

4.5 int * a
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right!

4.6 int 12
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right!

4.7 int **
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right!

4.8 char *
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Correct!

4.9 invalid
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Functions in C are actually just pointers to a spot in the program where some code exists. Just like you've been creating pointers to structs, strings, and arrays, you can point a pointer at a function too. The main use for this is to pass "callbacks" to other functions, or to simulate classes and objects. In this exercise we'll do some callbacks, and in the next one we'll make a simple object system.
The format of a function pointer goes like this:

int (*POINTER_NAME)(int a, int b)

So the answer is actually int(*)(int, char**). All you needed to do was evaluate the data types of the &main. Since & returns the memory address, main's data type would change from int to int*. Since argc and argv are just the parameters, they would remain the same data types. So the final answer is int(*)(int, char**).

4.10 unsigned long int 8
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right!

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PLEASE NOTE: Add explanations next time.

21 changes: 21 additions & 0 deletions reverse.c
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/*Christopher Liu */
#include <stdio.h>
#include <string.h>

int main()
{
printf("String: ");
char string[100];
fgets(string, 100, stdin);

char *stringPtr = string;
char *end_stringPtr = &string[strlen(string)];

while (*stringPtr != *end_stringPtr)
{
printf("%c", *(--end_stringPtr));
}
printf("\n");

return 0;
}