CSCB09: Software Tools and Systems Programming. Announcement. The address space. Memory model

Transcription

1 CSCB9: Sotware Tools and Systems Programming Announcement I needed to move my oice hour today to 11am. Bianca Schroeder bianca@cs.toronto.edu IC 46 Memory model Logical address I you know such people tell them to come. The address space The memory or a process (a running program) is called its address space Memory is just a sequence o bytes A memory location (a byte) is identiied by an address. I ll be attending an event at pm in IC1 or students interesting in applying or subject PoSts in CS, math, stats. Logical address Data Space or global variables and variables declared as static Dynamic Data Dynamic Space or dynamically allocated structures Unused Logical Address Space Space or variables created in unction calls: a unction s parameters and a unction s local variables

2 An example x Arrays x[] x int x = 1; int (int p, int q) { int j = ; return p * q + j; } int () { int i = x; y = (i, x); return ; } x8494 x x84944 y xa j xa4 p xa8 q xac i int x[]; or (i = ; i < ; i++) { x[i] = i*i; } x[1] x[2] x[] x[4] Arrays in C are a contiguous chunk o memory that contain a list o items o the same type.? x x x c x x I an array o ints contains 1 ints, then the array is 4 bytes. There is nothing extra. In particular, the size o the array is not stored with the array. There is no runtime checking. So you can access x[ ] What does that mean? 6 Pointer Arithmetic The array access operator [ ] is really only a shorthand or pointer arithmetic + dereerence These are equivalent in C: a[i] == *(a + i) Why do pointers need a type, e.g. int* or char*? So or a[ ], the program will happily try to access contents at address *(a ) l Behaviour o exceeding array bounds is undeined à à à program might appear to work program might crash (segmentation ault) program might do something apparently random 7 2

3 Why do pointers need a type, e.g. int* or char*? Why do pointers need a type, e.g. int* or char*? int i = ; int a[4] = {, 1, 2, }; int *p; p = a; or(i = ; i < 4; i++) { print("%d\n", *(p + i)); } (*p) == a[] *(p + 1) == a[1] *(p + 2) == a[2] *(p + ) == a[] 1 2 x114 x1144 x1148 x114c Pointer arithmetic respects the type o the pointer. E.g., int *p; *(p+1) ; really adds 4 to p (*p) == a[] *(p + 1) == a[1] *(p + 2) == a[2] *(p + ) == a[] 1 2 x114 x1144 x1148 x114c Hint: Why does adding 1 to p move it to the next spot or an int, when an int is 4 bytes? 9 C knows the size o what is being pointed at rom the type o the pointer. 1 Questions about pointers beore we start with page 1 o the Pointer Worksheet? 11 12

4 Passing unction x Let s see why lie() on the worksheet did not work as expected? int x = 1; int (int p, int q) { int j = ; p = ; return p * q + j; } int () { int i = x; y = (i, i); return ; } I () were to modiy p or q, will that change the value o s int i? x8494 x x84944 y xa j xa4 p xa8 q xac i Passing unction x Passing unction x int x = 1; int (int p, int q) { int j = ; x = ; return p * q + j; } int () { int i = x; y = (i, i); return ; } x8494 x x84944 y xa j xa4 p xa8 q int x = 1; void (int *p, int q) { *p = ; } int () { int i = x; (&i, i); return ; } When calling (): x8494 x x84944 y xa4 p xa8 q 1 xab 1 xac i 1 xac i 1 I ()were to modiy x or y would this change be permanent?

5 Passing unction x int x = 1; void (int *p, int q) { *p = ; } int () { int i = x; (&i, i); return ; } x8494 x x84944 y 1 With this in mind try to work through questions -6 on the Pointer Worksheet Ater () returns: xa4 p xa8 q xab 1 xac i Strings Strings are not a built-in type A string is an array o chars terminated by null character ( \ ). Beore we look at the Command-line Worksheet, a ew words about strings.. Initializing a string: char course_name[8] = { c, s, c, b,, 9, h, \ }; Or more conveniently: char course_name[8] = cscb9h ; Now you can do all the usual array operations, e.g. course_name[] = z ; Other ways to initialize the string: char course_name[8] = cscb9h ; 19 char course_name[2] = cscb9h ; // OUCH 2

6 Strings Ater initializing a string: char course_name[8] = { c, s, c, b,, 9, h, \ }; What is the type o course_name? char * What type would an array o strings be? char ** So what is argv in int (int argc, char **argv)? Now try to work through Q2- on the command-line argument sheet An array o strings That s it or today 6