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19 changes: 19 additions & 0 deletions Assignment7.txt
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Clarke Littlejohn

1) if i recall correctly ++*P moves the pointer then gets the values, *p++ get the value in the point the increments it
and *++p is not an apoerator that works.
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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).


2) I am confused by what this si asking it will go by what it is prendence so to will go from right to left or left to right based on the highest order
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The simple answer to this question is neither. C doesn’t always evaluate left-to-right or right-to-left. Generally, function calls are evaluated first, followed by complex expressions and then simple expressions.


3) you are working directly with the value so it is always modifing the number.
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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)string/char array/
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close! we're looking for the specific data type, so it's char*

.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!

.3)invalid
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This is perfectly valid! '\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)pointer this gets the address of the array
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the answer is 10 because it's just pointing to the first element of the array!

.5)address this gets the address
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This actually returns the type int*. Recall that a[0] would be of type int and the ampersand just returns the memory address. So &a[0] would return the pointer to the address (since a pointer is basically an address, which is why all pointers are 8 bytes no matter what they're pointing to), which would be of type int*.

.6)int this derefences the value
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right! but you need to evaluate the integer expression. Since a is a pointer to the first element of the array, the increment by 2 would move the pointer two spots over to 12. so 12 is the answer.

.7)pointer address p is pointer and ampersand gets the address
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right so the answer is int** for the a similar reason as number 5. The address of p returns a pointer to the address, but since p is of type int*, that means it will return the memory address of where the memory address is stored - in other words, a pointer to a pointer.

.8) this seems like incorrect syntax so it is invalid.
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It's actually``char*. This is because the * in ++argv dereferences the char* argv, leading to char **. The incrementing actually does nothing. In the same way that int x = 1; ++x would still be an int!`

.9)invalid is invalid as main is a function and this should not work.
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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**).

.10) it will return a value that is is 6
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Like I said in class, this is just a definition question, so the answer is 8 because pointers always allocate 8 bytes.

59 changes: 59 additions & 0 deletions catcmp.c
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#include<stdio.h>

int strcmp(char*,char*);
char* strct(char*,char*);



int main(){

printf("\n");
char *a = {"Hello"};
char *b = {"World"};
int aa=strcmp(a,b);
printf("%d",aa);
printf("\n");
return 0;
}

char * strct(char *one, char *two){
char a[(sizeof(one)+sizeof(two)-4)];
int i=0;
while(*one!='\0'){
a[i]=*one;
one++;
i++;
}
while(*two!='\0'){
a[i]=*two;
two++;
i++;
}
a[i]='\0';
one=a;
return one;
}


int strcmp(char* one,char* two){

while(*one==*two){
one++;
two++;
}
return ((int)(*one)-(int)(*two));
}














37 changes: 37 additions & 0 deletions reverse.c
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//Clarke Littlejohn
//reverse.c
// i had no idea how to use the second pointer, it seems useless to me as i just used a tracker it see how far
//away the end was from the start

#include<stdio.h>


void reverse(char[]);

int main(){

char str1[100];

printf("Enter in a word.");
fgets(str1,sizeof(str1),stdin);
reverse(str1);

}


void reverse(char str1[]){

int tracker = 0;
char* pStr=str1;
char* pStr2;
char start = *pStr;
while(*pStr!='\n'){pStr++;tracker++; }
*pStr='\0';
printf("%s\n",str1);
for(;tracker>0;tracker--){
pStr--;
printf("%c",*pStr);
}


}