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August 9, 2024 14:32
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the c programming language book, exercise 3-5
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| #include <stdio.h> | |
| /* | |
| itob(n,s,b) converts the integer n into a base | |
| b character representation in the string s. | |
| b could be up to 36. after that we run out of letters. | |
| */ | |
| void itob(int n, char h[], int b); | |
| void reverse(char s[], int len); | |
| void main() { | |
| char h[100]; | |
| itob(3552353, h, 36); | |
| printf("%s\n", h); | |
| } | |
| void itob(int n, char h[], int b) { | |
| int i, rem, len; | |
| i = 0; | |
| while (n != 0) { | |
| rem = n % b; | |
| if(rem <= 9) { | |
| h[i++] = rem + '0'; | |
| }else { | |
| h[i++] = rem + '0' + 7; | |
| } | |
| n /= b; | |
| } | |
| h[i] = '\0'; | |
| reverse(h, i); | |
| } | |
| void reverse(char s[], int len) { | |
| int c, i, j; | |
| for ( i = 0, j = len - 1; i < j; i++, j--) { | |
| c = s[i]; | |
| s[i] = s[j]; | |
| s[j] = c; | |
| } | |
| } |
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