Computer Systems C S Cynthia Lee
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1 Computer Systems C S Cynthia Lee
2 2 Topics ARRAYS AND POINTERS IN C: This will solidify and deepen your understanding of C strings They re just pointers/arrays of char! Pointers to arrays, Pointers -vs- arrays Pointer arithmetic -vs- bracket notation When is ? Arrays passed by reference memcpy / memmove
3 Test question: WHAT ARE YOUR CORE VALUES OR DEFINITIONS OF SUCCESS? (LIST AS MANY AS YOU LIKE; EX: LOYALTY, INDEPENDENCE, CREATIVITY, SPIRITUALITY/RELIGION,...)
4 Pointers in C REMEMBER: JUST A MEMORY ADDRESS
5 Pointers to Pointers // file: ptr_ex.c #include<stdio.h> #include<stdlib.h> Address Value int main(int argc, char **argv) { int x = 7; int *xptr = &x; // pointer to x printf("x: %d\n",x); printf("address of x: %p\n",xptr); printf("x via dereferencing xptr: %d\n",*xptr); } return 0; 0x7ffee40f1494 0x7ffee40f1490 0x7ffee40f148c 7 x $./ptr_ex x: 7 address of x: 0x7ffee40f148c address of xptr: 0x7ffee40f1480 x via dereferencing xptr: 7 xptr 0x7ffee40f148c 0x7ffee40f1488 0x7ffee40f1480 0x7ffee40f148c xptr
6 Pointers to Pointers // file: ptrptr_ex.c #include<stdio.h> #include<stdlib.h> Address Value int main(int argc, char **argv) { int x = 7; int *xptr = &x; // pointer to x int **xptrptr = &xptr; // pointer to xptr printf("x: %d\n",x); printf("address of x: %p\n",xptr); printf("address of xptr: %p\n",xptrptr); printf("address of x via dereferencing xptrptr: %p\n",*xptrptr); printf("x via double-dereferencing xptrptr: %d\n",**xptrptr); 0x7ffee40f1494 0x7ffee40f1490 0x7ffee40f148c 7 x } return 0; 0x7ffee40f1488 $./ptrptr_ex xptrptr x: 7 0x7ffee40f1480 address of x: 0x7ffee40f148c address of xptr: 0x7ffee40f1480 address of x via dereferencing xptrptr: 0x7ffee40f148c x via double-dereferencing xptrptr: 7 0x7ffee40f1480 0x7ffee40f148c xptr
7 Question: what does the following code print out? // file: ptrptr_mystery.c #include<stdio.h> #include<stdlib.h> int main(int argc, char **argv) { char *str = "CS107"; char **strptr = &str; } char mystery = **strptr; printf("mystery: %c\n",mystery); return 0;
8 Question: what does the following code print out? // file: ptrptr_mystery.c #include<stdio.h> #include<stdlib.h> int main(int argc, char **argv) { char *str = "CS107"; char **strptr = &str; Answer: C A single dereference *strptr produces the address of str. If we were to dereference str: } char mystery = **strptr; printf("mystery: %c\n",mystery); return 0; *str this would give us the first character of str, or 'C'.
9 Arrays in C Arrays in C are contiguous blocks of memory with a fixed length. Access elements in an array using either bracket notation arr[2] OR, by dereferencing an offset from the beginning of the array *(arr + 2) Although arrays sometimes behave like pointers, they are distinct and should not be confused with pointers. Address 0x7ffeea3c9498 0x7ffeea3c9494 0x7ffeea3c9490 0x7ffeea3c948c 0x7ffeea3c9488 0x7ffeea3c9484 Value
10 Arrays in C // file: arrays_not_pointers.c #include<stdio.h> #include<stdlib.h> int main(int argc, char **argv) { int arr[3] = {1,2,3}; int a = 4; arr = &a; // produces compile error return 0; } You cannot assign a different value to an array, because it isn't a pointer. $ make arrays_not_pointers gcc -g -O0 -std=gnu99 -Wall $warnflags arrays_not_pointers.c -o arrays_not_pointers arrays_not_pointers.c:9:9: error: array type 'int [3]' is not assignable arr = &a; // produces compile error ~~~ ^ 1 error generated. make: *** [arrays_not_pointers] Error 1
11 Arrays in C // file: arr_to_ptr.c #include<stdio.h> #include<stdlib.h> int main(int argc, char **argv) { int arr[3] = {1,2,3}; int *arrptr = arr; arrptr = &arr[0]; // also works printf("%p\n",arrptr); printf("%p\n",arr); printf("%p\n",&arr[0]); printf("%p\n",&arr); // also works, but try to avoid this return 0; } $./assign_arr_to_ptr 0x7ffee363d60c 0x7ffee363d60c 0x7ffee363d60c 0x7ffee363d60c
12 Arrays in C int arr[] = {8,2,7,14,-5,42}; int *arrptr = arr; // arr now points to the 8 arrptr 0x7ffeea3c9484 Address Value 0x7ffeea3c x7ffeea3c x7ffeea3c x7ffeea3c948c 7 0x7ffeea3c x7ffeea3c arr
13 Arrays in C int arr[] = {8,2,7,14,-5,42}; int *arrptr = arr; // arr now points to the 8 arrptr++; // arrptr now points to the 2 arrptr? Address Value x7ffeea3c arr
14 Question: What is the value of arrptr now? (A) 0x7ffeea3c9485 (B) 0x7ffeea3c9486 (C) 0x7ffeea3c9488 (D) 0x7ffeea3c948C int arr[] = {8,2,7,14,-5,42}; int *arrptr = arr; // arr now points to the 8 arrptr++; // arrptr now points to the 2 arrptr? Address Value x7ffeea3c arr
15 Arrays in C Answer: (C) Moving to the next int value is 4bytes away note that ++ operator needs to know that arrptr points to ints (and how big one int is)! int arr[] = {8,2,7,14,-5,42}; int *arrptr = arr; // arr now points to the 8 arrptr++; // arrptr now points to the 2 arrptr 0x7ffeea3c9488 Address Value 0x7ffeea3c x7ffeea3c x7ffeea3c x7ffeea3c948c 7 0x7ffeea3c x7ffeea3c arr
16 memcpy, memmove Used to copy contents of memory, starting at an address (pointer) void *memcpy(void *dest, const void *src, size_t n); void *memmove(void *dest, const void *src, size_t n); Memcpy only works if the source and destination blocks of memory do NOT overlap! If they do overlap (say, moving a block of 10 bytes just 3 bytes forward), then use memmove.
17 void* Void* is a type for holding an address, but when we don t know (don t specify) the size of the box at the end of the arrow void *memcpy(void *dest, const void *src, size_t n); void *memmove(void *dest, const void *src, size_t n); Important: what happens in part 2 of executing a dereference * operation on a void*? Hmm.
18 void* Void* is a type for holding an address, but when we don t know (don t specify) the size of the box at the end of the arrow void *memcpy(void *dest, const void *src, size_t n); void *memmove(void *dest, const void *src, size_t n); Important: what happens in part 2 of executing a dereference * operation on a void*? Hmm. You can t dereference a void*! (syntax error)
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