Dynamic Memory CMSC 104 Spring 2014, Section 02, Lecture 24 Jason Tang

Transcription

1 Dynamic Memory CMSC 104 Spring 2014, Section 02, Lecture 24 Jason Tang

2 Topics Memory Lifetime Dynamic Memory Allocation Dynamic Memory Deallocation

3 Variable Lifetime (Review) Lifetime refers to when the computer reserves memory to the variable and when the computer will re-use that memory afterwards A local variable s lifetime is the same as its scope Recall that every variable refers to a unique memory address

4 Why Use Dynamic Memory? Thus far, all strings have relied upon arrays An array s size is determined at array declaration and cannot be changed afterwards (so-called static allocation ) What if the array s size cannot be known ahead of time? Example scenario: Write a function that takes a string and returns a duplicate of it (a strdup() function)

5 Flawed strdup() Implementations char * bad_strdup(char *orig) { int len = strlen(orig); What s wrong char dup[len + 1]; with this? strcpy(dup, orig); return dup; char dup[????]; char * bad_strdup(char *orig) { int len = strlen(orig); What s wrong strcpy(dup, orig); with this? return dup;

6 malloc() Solution is to determine and allocate just the right amount of memory at runtime Use malloc() function to do this Takes an argument that specifies number of bytes to reserve Returns a pointer to the start of reserved bytes

7 Using malloc() #include <stdlib.h> #include <stdio.h> int main(void) { char *a = malloc(5); *(a + 0) = 'U'; *(a + 1) = 'M'; *(a + 2) = 'B'; *(a + 3) = 'C'; *(a + 4) = '\0'; printf("%s\n", a); return 0;

8 Using malloc() #include <stdlib.h> Need to #include <stdlib.h> #include <stdio.h> when calling malloc() int main(void) { char *a = malloc(5); *(a + 0) = 'U'; *(a + 1) = 'M'; *(a + 2) = 'B'; *(a + 3) = 'C'; *(a + 4) = '\0'; printf("%s\n", a); return 0; Bytes returned by malloc() still need to be initialized

9 Caveats of malloc() Memory that has been reserved by malloc() stays reserved until told otherwise Either implicitly when program terminates, or Explicitly (see next slide) That memory s scope and lifetime is from the invocation of malloc() until it is no longer reserved

10 free() Call free(), also declared in stdlib.h, to release memory previously reserved by a call to malloc() Takes as a parameter the same address that was returned by malloc() Generally speaking for every malloc(), there needs to be exactly one (and only one!) matching free()

11 Using free() #include <stdlib.h> #include <stdio.h> int main(void) { char *a = malloc(5); *(a + 0) = 'U'; *(a + 1) = 'M'; *(a + 2) = 'B'; *(a + 3) = 'C'; *(a + 4) = '\0'; printf("%s\n", a); free(a); return 0;

12 Caveats of free() Can only free() something that was returned by malloc() or some other memory allocation function free() must be against the exact same address that was returned by malloc() (or by some other allocater ) free() of a non-pointer or of an address not returned by malloc() will crash program This is also a segmentation fault

13 Crashing Program #1: Freeing a Non-Pointer #include <stdlib.h> #include <stdio.h> int main(void) { char a[] = "UMBC"; printf("%s\n", a); free(a); return 0;

14 Crashing Program #2: Not Freeing Exact Same Address #include <stdlib.h> #include <stdio.h> int main(void) { char *a = malloc(3); *(a + 0) = 'U'; *(a + 1) = 'M'; *(a + 2) = '\0'; printf("%s\n", a); free(a+1); return 0;

15 Crashing Program #3: Freeing a Non-Allocated Pointer #include <stdlib.h> #include <stdio.h> int main(void) { char a[] = "UMBC"; char *b = &a[0]; printf("%s\n", a); free(b); return 0;

16 Crashing Program #4: Doublefree() #include <stdlib.h> #include <stdio.h> int main(void) { char *a = malloc(3); *(a + 0) = 'U'; *(a + 1) = 'M'; *(a + 2) = '\0'; printf("%s\n", a); free(a); printf("freeing again\n"); free(a); return 0;

17 Crashing Program #4: Doublefree() #include <stdlib.h> #include <stdio.h> int main(void) { char *a = malloc(3); *(a + 0) = 'U'; *(a + 1) = 'M'; *(a + 2) = '\0'; printf("%s\n", a); free(a); printf("freeing again\n"); free(a); return 0; Once free()d, that memory block may not be accessed again This includes calling free() on pointer a second (or third or fourth) time

18 Working homemade strdup() with malloc() char * good_strdup(char *orig) { int len = strlen(orig); char *dup = malloc(len + 1); strcpy(dup, orig); return dup; This version of strdup() uses dynamic memory to reserve just the right amount of memory Don t forget to add 1 so as to hold the trailing \0

19 Rest of strdup() example #include <stdio.h> #include <stdlib.h> #include <string.h> char * good_strdup(char *orig); int main(void) { char s[] = "UMBC"; char *t = good_strdup(s); *(t + 0) = *(t + 1) = 'C'; printf("s = %s, t = %s\n", s, t); free(t); return 0;

20 strdup() This is a common design pattern: get the length of a string (strlen()), allocate space (malloc()), then copy the string into the returned memory (strcpy()) Built-in C library already has a function that does this, strdup() Declared in string.h like the other string functions strdup() is a memory allocator, so its returned

21 Cleaned Up Example #include <stdio.h> #include <stdlib.h> #include <string.h> int main(void) { char s[] = "UMBC"; char *t = strdup(s); *(t + 0) = *(t + 1) = 'C'; printf("s = %s, t = %s\n", s, t); free(t); return 0;

22 Cleaned Up Example #include <stdio.h> #include <stdlib.h> #include <string.h> int main(void) { char s[] = "UMBC"; char *t = strdup(s); *(t + 0) = *(t + 1) = 'C'; printf("s = %s, t = %s\n", s, t); free(t); return 0; strdup() is an allocator, so don t forget to free() the memory afterwards

23 Updated Variable Qualifiers Type Visibility Lifetime Storage Class scalar function local duration of function static allocation array global duration of program dynamic allocation pointer complex You are now ready for CMSC 201!