#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#include <unistd.h>
#include <errno.h>
int c, n, E, b, s, v, t, opt, valid = 0;
char current = '\0';
char previous = '\0';
FILE *fp;
/* -n numbers lines
* -E appends a dollar sign to line ends
* -b numbers only non-blank lines
* -s squeezes multiple blank lines down to 1
* -v displays control chars, excluding tab
* -t includes tab in the above
* -e is the same as -E and -v
*/
int setFlags(int argc, char *argv[]) {
int op;
while ((op = getopt(argc, argv, "nEbsvte")) != -1) {
switch (op) {
case 'n': {
n = 1;
break;
} case 'E': {
E = 1;
break;
} case 'b': {
b = 1;
break;
} case 's': {
s = 1;
break;
} case 'v': {
v = 1;
break;
} case 't': {
t = 1;
break;
} case 'e': {
E = 1;
v = 1;
break;
} case '?': {
//fprintf(stderr, "Option `-%c` is not valid.\n", optopt);
return EXIT_FAILURE;
} default: {
abort();
}
}
}
opt = optind;
if(n == 1) {
b = 0;
}
return EXIT_SUCCESS;
}
int checkFile(char *path) {
if (access(path, R_OK) == 0) {
return EXIT_SUCCESS;
} else {
fprintf(stderr, "cat: %s: %s\n", argv[i], strerror(errno));
errno = 0;
return EXIT_FAILURE;
}
}
int doPrint(char *path) {
if (strcmp(path, "stdin") == 0) {
fp = stdin;
} else {
if (checkFile(path) == 1) {
return EXIT_FAILURE;
} else {
fp = fopen(path, "r");
}
}
while ((c = fgetc(fp)) != EOF) {
putchar(c);
}
fclose(fp);
return EXIT_SUCCESS;
}
int main (int argc, char *argv[]) {
if (setFlags(argc, argv) == 1) {
fprintf(stderr, "The program has terminated with an error.\n"
"An invalid option was specified.\n");
return EXIT_FAILURE;
} else {
if ((argc - opt) == 0) {
doPrint("stdin");
} else {
for(int i = opt; i < argc; i++) {
doPrint(argv[i]);
}
}
}
}
我遇到了一个非常疯狂的错误,我的程序在完成写入文件内容之前在 checkFile 中输出错误行(总是在结束前进行一次聊天)。
它让我发疯,无论我将那段代码移动到哪里,它都无法按预期工作。
我敢肯定答案可能是微不足道的,但它让我难住了。我什至在输出完成之前投入了睡眠和其他各种事情,它会抛出错误,然后睡眠,然后打印最后一个字符。
有什么帮助吗?