Programming Review: Functions, pointers and strings

Pointers int nums[] = {1, 2, 3}; char str[] = "moshe"; int* q = nums; char* p = str; q num m o s h e \0 p str

Pointers int nums[] = {1, 2, 3}; char str[] = "moshe"; int* q = nums; char* p = str; (q+1) (p+3) m o s h e \0 str

Pointers p[i] *(p+i) m o s h e \0 q p q[1]q[2]q[0] p[1]p[4] p[0] *p

Pointers and Functions If we want to change a variable inside a function, we must pass it a pointer to the variable (its address) The function will “fill” this address with the right value Example: Swap void swap(int *x, int *y) { int temp = *x; *x = *y; *y = temp; }

Pointers and Functions Variables that are defined inside the function “die” when the function ends!!! char* func() { char str[LENGTH + 1];... return str; } str doesn’t exist outside the function’s body

1.What is wrong here? int main() { int n = 3; multBy3(n); printf(“n=%d”,n); } void multBy3(int n) { int num = n; num *= 3; n = num; }

2.What is wrong here? int main() { int n = 3; multBy3(&n); printf(“n=%d”,n); } void multBy3(int *n) { int num = n; num *= 3; n = num; }

3.What is wrong here? int main() { int n = 3; multBy3(&n); printf(“n=%d”,n); } void multBy3(int *n) { int num = *n; num *= 3; n = num; }

4.What is wrong here? int main() { int n = 3; multBy3(&n); printf(“n=%d”,n); } void multBy3(int *n) { int num = *n; num *= 3; n = # }

References 1. Define a variable in the main 2. Pass its address to the function 3. The function fills the address with a value 4. The main can use it as a normal variable int main() { int num; }

References 1. Define a variable in the main 2. Pass its address to the function 3. The function fills the address with a value 4. The main can use it as a normal variable Int main() { int num; multBy3(&num); }

References 1. Define a variable in the main 2. Pass its address to the function 3. The function fills the address with a value 4. The main can use it as a normal variable void multBy3(int *n) { (*n) *=3; } Int main() { int num; multBy3(&num); }

References 1. Define a variable in the main 2. Pass its address to the function 3. The function fills the address with a value 4. The main can use it as a normal variable Int main() { int num; multBy3(&num); printf(“num=%d”,num); }

Exercise with pointers and strings Implement the following function: char* str_any(char *str1, char *str2);  Input – two strings str1, str2  Output – pointer to the first occurrence in str1 of any of the characters contained in str2

Exercise (cont.) Write a program that accepts a string from the user and replaces all punctuation signs (,.;:!? ) with spaces

Solution (str_any.c) char* str_any(char* str1, char* str2) { while (*str1 != '\0') { if (strchr(str2, *str1) != NULL) { return str1; } ++str1; } return NULL; }

Solution int main(void) { char* punc = ".,;:!?"; char s[MAX_LENGTH + 1]; char *p; printf("Please enter a line of text\n"); scanf("%100s", s); for (p = str_any(s, punc); p != NULL; p = str_any(p + 1, punc)) { *p = ' '; } printf("Resulting string is:\n%s\n", s); return 0; }

Command line arguments Command line arguments are arguments for the main function  Recall that main is basically a function  It can receive arguments like other functions  The ‘calling function’ in this case is the operating system, or another program

‘main’ prototype When we want main to accept command line arguments, we must define it like this  argc holds the number of arguments that were entered by the caller  argv is an array of pointers to char – an array of strings – holding the text values of the arguments The first argument is always the program’s name int main(int argc, char* argv[])

‘main’ prototype int main(int argc, char* argv[]) argc : 3 argv : \0 p r o g n a m e \0 t e x t \0

Example /* This program displays its command-line arguments */ #include int main(int argc, char *argv[]) { int i; printf("The program's command line arguments are: \n"); for (i = 0; i < argc; ++i) { printf("%s\n", argv[i]); } return 0; }

Specifying the arguments In Visual Studio: Project  Settings  Debug, in the ‘program arguments’ field

Specifying the arguments We can also specify the arguments directly, by using the Windows console (Start  Run…, then type ‘cmd’ and drag the executable into the window. Then type the arguments and )

Helper functions – atoi/atof Command line arguments are received in the form of strings These functions are used when we want to transform them into numbers For example – atof(“13.5”) returns the number Must #include int atoi(char s[]); double atof(char s[]);

Exercise Write a program that accepts two numbers as command line arguments, representing a rectangle’s height and width (as floating- point numbers). The program should display the rectangle’s area and perimeter

Solution (args_rectangle.c) int main(int argc, char* argv[]) { double width, height; if (argc != 3) { printf("Wrong number of arguments!\n"); return 1; } width = atof(argv[1]); height = atof(argv[2]); printf("The rectangle's area is %g\n", width * height); printf("The rectangle's perimeter is %g\n", 2 * (width + height)); return 0; }