CSC 373 Sections 501, 510 Winter 2015 C Examples

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1 CSC 373 Sections 501, 510 Winter 2015 C Examples These are posted on condor. These notes do not have any #include directives. The.c files should compile as they are 1. Simple hello: 1_hello.c 2. Illustrates no bounds checking: 2_bounds.c // incorrect; the limit on i should of course be i < 5. But the compiler will not detect it, and not // even the runtime environment will detect it until a sudden program crash int x[5] = {1, 2, 3, 4, 5; for (i = 0 ;; i++) printf("%#x, %d\n", x+i, x[i]); Complete output is in 2.txt Portion of output: at the beginning $./2 0x24f7dd50, 1 0x24f7dd54, 2 0x24f7dd58, 3 0x24f7dd5c, 4 0x24f7dd60, 5 0x24f7dd64, x24f7dd68, 0 0x24f7dd6c, 7 beyond end of array; contains garbage from here on At the end: 0x173b5ff4, x173b5ff8, 0 0x173b5ffc, 0 Segmentation fault eventually, code hits a memory address which is not accessible to the process

2 3. Correct usage of arrays other than C strings: 3_arraysyntax.c void print(int x[], int l); int x[10]; for (i=0; i<10; i++) x[i] = i*5; print(x, 10); return 0; // note that you need to pass the length of the array as a separate parameter void print(int x[], int l) { for (i=0; i<l; i++) printf("%d ", x[i]); printf("\n"); Output: Same output, but pointer syntax: 4_pointersyntax.c // don t have alter main void print(int *x, int l) { for (i=0; i<l; i++) printf("%d ", *(x+i)); printf("\n"); Same output

3 5. Examples of C strings: 5_strings.c void my_print(char [ ]); // array syntax void my_print2(char *); // pointer syntax char word[20]; printf("enter a word\n"); scanf("%s", word); printf("you entered\n"); my_print(word); printf("\n"); my_print2(word); printf("\n"); return 0; // no need to pass array length with strings. Also, you could produce the same output with // %s, but I wrote it this way to reflect the underlying representation of cstrings. void my_print(char c[ ]) { for (i=0; c[i]!= '\0'; i++) printf("%c", c[i]); // again, you can do the same thing with pointer syntax void my_print2(char *c) { for (i=0; *(c+i)!= '\0'; i++) printf("%c", *(c+i)); Repeats whatever string the user entered, twice.

4 6. Illustrates pointers passed as parameters int twice(int *); int main( ) { int y; int *x = &y; y = 3; printf("before call to twice: y = %d\n", y); int d = twice(x); printf("after call to twice. twice returns %d, y = %d\n", d, y); return 0; int twice(int *x) { *x = *x * 2; return *x; Output: Before call to twice: y = 3 After call to twice. twice returns 6, y = 6 7. Illustrates pointer values as hexadecimal numbers: 7_pointers.c int main() { int a[5] = {5, 10, 15, 20, 25; int *p = a; printf("a = %#x\n", a); // print as hexadecimal printf("p = %#x %u %d %i\n", p, p, p, p); // print as hex, decimal printf("a[0] = %d\n", a[0]); printf("a[1] = %d\n", a[1]); printf("a[2] = %d\n", a[2]); printf("*p = %d\n", *p); printf("*(p+1) = %d\n", *(p+1)); printf("*(p+2) = %d\n", *(p+2)); printf("p[0] = %d\n", p[0]); printf("p[1] = %d\n", p[1]); printf("p[2] = %d\n", p[2]);

5 printf("*a = %d\n", *a); printf("*(a+1) = %d\n", *(a+1)); printf("*(a+2) = %d\n", *(a+2)); Output: hex numbers may be machine-dependent [steveadmin@slytinen373 linux_c]$./7 a = 0xa7ff2180 p = 0xa7ff a[0] = 5 a[1] = 10 a[2] = 15 *p = 5 *(p+1) = 10 *(p+2) = 15 p[0] = 5 p[1] = 10 p[2] = 15 *a = 5 *(a+1) = 10 *(a+2) = Sizeof operator: 8_sizeof.c #include <stdio.h> void fn(int x[ ]) { printf("in fn\nint array: %d\n", sizeof x); int a; long b; short c; float d; double e; char f; int *g; int h[20]; printf("int: %d\nlong: %d\nshort: %d\nfloat: %d\n", sizeof a, sizeof b, sizeof c, sizeof d);

6 printf("double: %d\nchar: %d\nint *: %d\nint array: %d\n", sizeof e, sizeof f, sizeof g, sizeof h); fn(h); return 0; Ouput on 64-bit platform int: 4 long: 8 short: 2 float: 4 double: 8 char: 1 int *: 8 int array: 80 In fn int array: 8 9. Compare ASCII codes with their character representations: 9_ascii.c for (i = 0; i < 128; i++) printf("%c: %d\n", i, i); Run for yourself 10. Compare different formats: 10_format.c printf ("Characters: %c %c \n", 'a', 65); printf ("Characters: %d %d \n", 'a', 65); printf ("Decimals: %d %ld\n", 1977, ); printf ("Preceding with blanks: %10d \n", 1977); printf ("Preceding with zeros: %010d \n", 1977); printf ("Some different radixes: %d %x %o %#x %#o \n", 100, 100, 100, 100, 100); printf ("floats: %4.2f %+.0e %E \n", , , ); printf ("Width trick: %*d \n", 5, 10); printf ("%s \n", "A string"); int x;

7 float y; printf("enter a decimal and a floating point number\n"); scanf("%d %f", &x, &y); printf("you entered %d %f\n", x, y); I/O (All is output except which is user input) Characters: a A Characters: Decimals: Preceding with blanks: 1977 Preceding with zeros: Some different radixes: x floats: e E+00 Width trick: 10 A string Enter a decimal and a floating point number You entered Illustrate bit-operators: 11_bitops.c // shows the effect of various bit operations #include <stdio.h> void print_binary(int x) { int size = (sizeof x) * 8; // makes this machine-independent // 2's complement if (x<0) printf("1"); else printf("0"); int mask = 1; for (i=1; i<size-1; i++) mask = mask << 1; for (i=1; i<size; i++) { if (x&mask) printf("1"); else printf("0"); mask = mask >> 1;

8 int x,y; unsigned int z; printf("enter 2 numbers\n"); scanf("%d %d", &x, &y); z = x; printf("x in binary is "); print_binary(x); printf("\ny in binary is "); print_binary(y); printf("\n\nx & y = "); print_binary(x&y); printf("\nx y = "); print_binary(x y); printf("\nx ^ y = "); print_binary(x^y); printf("\n\nx = "); print_binary(x); printf("\n~x = "); print_binary(~x); printf("\nx<<1 = "); print_binary(x << 1); printf("\nx<<2 = "); Output (assuming input of 30 23) [steveadmin@slytinen373 linux_c]$./11 Enter 2 numbers x in binary is y in binary is x & y = x y = x ^ y = x = ~x = x<<1 = x<<2 = x>>3 =

9 x in decimal = 30 ~x in decimal = -31 x<<1 in decimal = 60 x<<2 in decimal = 120 x>>1 in decimal = 15 z = ~z = z<<1 = z<<2 = z>>3 = z in decimal = 30 ~z in decimal = ~z<<1 in decimal = 60 ~z<<2 in decimal = 120 ~z>>1 in decimal = 15