From e6b7015c5dd0431aaaf40256d1d2db3cda6769bb Mon Sep 17 00:00:00 2001 From: zlg Date: Tue, 23 Apr 2013 02:09:09 -0500 Subject: Solve Exercise 3-6: itoa (3 arg version) --- ch3/3-06_itoa3.c | 67 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 67 insertions(+) create mode 100644 ch3/3-06_itoa3.c diff --git a/ch3/3-06_itoa3.c b/ch3/3-06_itoa3.c new file mode 100644 index 0000000..3c28b0b --- /dev/null +++ b/ch3/3-06_itoa3.c @@ -0,0 +1,67 @@ +#include +#include +#include + +/* The C Programming Language: 2nd Edition + * + * Exercise 3-6: Write a version of itoa that accepts three arguments instead + * of two. The third argument is a minimum field width; the converted number + * must be padded with blanks on the left if necessary to make it wide enough. + * + * Answer: This is trivial. Process the number, then check the value of i. If + * it's less than the field width, add a blank until i is equal to the field + * width and terminate your string. + */ + +void reverse(char s[]) { + int c, i, j; + + for (i = 0, j = strlen(s)-1; i < j; i++, j--) { + c = s[i]; + s[i] = s[j]; + s[j] = c; + } +} + +void itoa(int n, char s[], unsigned fw) { + int i, sign, min; // Add 'min' for later use + + if ((sign = n) < 0) { + /* Detect this and add one so it can properly be made positive */ + if (n == INT_MIN) { + n += 1; + min = 1; + } + n = -n; + } + i = 0; + do { + /* If it's the first iteration of the loop and we've established + * that n == INT_MIN, we need to add one to the resulting string for it + * to be displayed properly. */ + if (i == 0 && min == 1) { + s[i++] = n % 10 + '1'; + } else { + s[i++] = n % 10 + '0'; + } + } while ((n /= 10) > 0); + if (sign < 0) { + s[i++] = '-'; + } + while (i < fw) { + s[i++] = ' '; + } + s[i] = '\0'; + reverse(s); +} + +int main() { + int i; + char foo[16] = ""; + int tests[5] = {-25, 409689, 8, -1000, 135}; + for (i = 0; i < 5; i++) { + itoa(tests[i], foo, 5); + printf("Pad to 5 spaces! '%s'\n", foo); + } + return 0; +} -- cgit v1.2.3-54-g00ecf