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Jack Rosen, Assignment 7#2
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,14 @@ | ||
| Jack Rosen | ||
| 1. ++*p increments the value at point p. *p++ increments p and then returns the value. *++p increments p and then returns the value as well. | ||
| 2. It is only guaranteed for certain operators. The first level of operators goes from left to right, the second goes from right to left, and the rest go from left to right. | ||
Contributor There was a problem hiding this comment. 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.The advantages of using pointers is that they are very powerful and help a person work with arrays. It is also a way to access a variable without having to call for it. | ||
Contributor There was a problem hiding this comment. Pointers are more efficient in handling arrays and data tables and can be used to return multiple values from a function via function arguments. They also allow references to functions and C to support dynamic memory management. Lastly, Pointers reduce length and complexity of programs! | ||
| 4.1 char * | ||
Contributor There was a problem hiding this comment. Please note that while you got this right, you didn't explain your answers! | ||
| 4.2 invalid | ||
Contributor There was a problem hiding this comment. 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 invalid | ||
Contributor There was a problem hiding this comment. 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. | ||
| 4.4 10 | ||
Contributor There was a problem hiding this comment. Yup! but again, please explain answers as the directions said. | ||
| 4.5 0 | ||
Contributor There was a problem hiding this comment.
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| 4.6 12 | ||
Contributor There was a problem hiding this comment. Right! but again, explain your answers! | ||
| 4.7 2 | ||
Contributor There was a problem hiding this comment.
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| 4.8 char | ||
Contributor There was a problem hiding this comment. Close, but not quite! You're right that it's char, but 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! | ||
| 4.9 0 | ||
Contributor There was a problem hiding this comment.
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| 4.10 6 | ||
Contributor There was a problem hiding this comment. sideof(str) is always 8 because all pointers are 8 bytes no matter what they're pointing to. | ||
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| #include <stdio.h> | ||
| #include <string.h> | ||
| void stringReversal(char *input) | ||
| { | ||
| char *ptrBegin = input, *ptrEnd = ptrBegin + strlen(input) - 1; | ||
| for (; ptrBegin <= ptrEnd; --ptrEnd) | ||
| { | ||
| if (*ptrEnd == '\n') | ||
| { | ||
| continue; | ||
| } | ||
| printf("%c", *ptrEnd); | ||
| } | ||
| } | ||
| int main() | ||
| { | ||
| printf("Write anything you want!\n"); | ||
| char input[1000] = {}; | ||
| int i; | ||
| fgets(input, sizeof(input), stdin); | ||
| for (i = 0; i < strlen(input); i++) | ||
| { | ||
| if (input[i] == '\n' || input[i] == '\0') | ||
| { | ||
| i--; | ||
| break; | ||
| } | ||
| } | ||
| stringReversal(input); | ||
| printf("\n"); | ||
| return 0; | ||
| } |
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,52 @@ | ||
| #include <stdio.h> | ||
| #include <string.h> | ||
| void concat(char *ptr1, char *ptr2) | ||
| { | ||
| char *ptr3 = ptr1; | ||
| while (*ptr3 != '\0') | ||
| { | ||
| ptr3++; | ||
| } | ||
| while (*ptr2 != '\0') | ||
| { | ||
| *ptr3 = *ptr2; | ||
| ptr3++; | ||
| ptr2++; | ||
| } | ||
| ptr3++; | ||
| *ptr3 = '\0'; | ||
| } | ||
| int compare(char *ptr1, char *ptr2) | ||
| { | ||
| int flag=0; | ||
| while(*ptr1!='\0' && *ptr2!='\0'){ | ||
| if(*ptr1 != *ptr2){ | ||
| flag=1; | ||
| break; | ||
| } | ||
| ptr1++; | ||
| ptr2++; | ||
| } | ||
| if (flag==0 && *ptr1=='\0' && *ptr2=='\0') | ||
| { | ||
| return 0; | ||
| } | ||
| else if (*ptr1 > *ptr2) | ||
| { | ||
| return 1; | ||
| } | ||
| else if (*ptr2 > *ptr1) | ||
| { | ||
| return -1; | ||
| } | ||
| } | ||
| int main() | ||
| { | ||
| char ptr1[12] = "Hello ", ptr2[] = "World"; | ||
| concat(ptr1, ptr2); | ||
| printf("%d\n", compare(ptr1,ptr2)); | ||
| printf("%s\n", ptr1); | ||
| return 0; | ||
| } |
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Hmm, not quite. 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).