Improvements to expseq help text
Getty Ritter
12 years ago
| 2 | 2 | #include <stdio.h> |
| 3 | 3 | #include <stdlib.h> |
| 4 | 4 | |
| 5 | #define HELPMSG "USAGE: expseq LAST\n or: expseq FIRST LAST\n or: expseq --help\n"\ | |
| 6 | "Print numbers 2^FIRST to 2^LAST in exponential steps.\n\n"\ | |
| 7 | "If FIRST is omitted, it defaults to 0.\n" | |
| 8 | ||
| 5 | 9 | int main(int argc, char* argv[]) |
| 6 | 10 | { |
| 7 |
int fstnum = 0, lstnum |
|
| 11 | int fstnum = 0, lstnum, i; | |
| 8 | 12 | if (argc == 2) { |
| 9 | char* endptr = argv[1]; | |
| 10 | lstnum = strtol(argv[1], &endptr, 10); | |
| 11 | if (!endptr) { | |
| 12 | fprintf(stderr, "%s: non-numeric arg: %s\n", argv[0], argv[1]); | |
| 13 | return 0; | |
| 14 | } | |
| 13 | if (strcmp(argv[1], "--help")==0 | |
| 14 | || strcmp(argv[1], "-h")==0) { | |
| 15 | fprintf(stderr, HELPMSG); | |
| 16 | return 1; | |
| 17 | } else { | |
| 18 | char* endptr = argv[1]; | |
| 19 | lstnum = strtol(argv[1], &endptr, 10); | |
| 20 | if (!endptr) { | |
| 21 | fprintf(stderr, "%s: non-numeric arg: %s\n", argv[0], argv[1]); | |
| 22 | return 1; | |
| 23 | } | |
| 24 | } | |
| 15 | 25 | } else if (argc == 3) { |
| 16 | 26 | char* endptr = argv[1]; |
| 17 | 27 | fstnum = strtol(argv[1], &endptr, 10); |
| 18 | 28 | if (!endptr) { |
| 19 | 29 | fprintf(stderr, "%s: non-numeric arg: %s\n", argv[0], argv[1]); |
| 20 |
return |
|
| 30 | return 1; | |
| 21 | 31 | } |
| 22 | 32 | endptr = argv[2]; |
| 23 | 33 | lstnum = strtol(argv[2], &endptr, 10); |
| 24 | 34 | if (!endptr) { |
| 25 | 35 | fprintf(stderr, "%s: non-numeric arg: %s\n", argv[0], argv[2]); |
| 26 |
return |
|
| 36 | return 1; | |
| 27 | 37 | } |
| 28 | 38 | } else { |
| 29 | 39 | fprintf(stderr, "%s: missing operand\n", argv[0]); |
| 30 |
return |
|
| 40 | return 1; | |
| 31 | 41 | } |
| 32 |
i |
|
| 42 | i = fstnum; | |
| 33 | 43 | while (i <= lstnum) { |
| 34 | 44 | printf("%d\n", (int) pow(2, (float) i)); |
| 35 | 45 | i++; |