我有以下代码,我想为简单的计算器语言创建一个简单的扫描程序。我正在使用fgetc从文件中获取字符。虽然,在某些地方我还需要检查所遵循的下一个字符。出于这个原因,我一直在使用++运算符,但它似乎无法正常工作。有人可以帮我解决我的问题。
例如,当我在我的文本文件中有:=时,它会打印冒号,然后显示一条错误消息:“;不能跟随:(冒号)。”,而它应该打印“assign”。
这是我的完整代码:
#include <regex.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/types.h>
int main(int argc, char** argv)
{
FILE *fp;
int c, check, reti_d, reti_l;
regex_t digit, letter;
reti_d = regcomp(&digit, "[0-9]", 0);
reti_l = regcomp(&letter, "[a-zA-Z]", 0);
if (reti_d || reti_l)
{
fprintf(stderr, "Couldn't compile the regular expression(s).\n");
exit(1);
}
if (argc != 2)
{
fprintf(stderr, "Usage: %s filename.txt\n", argv[0]);
exit(1);
}
if (!(fp = fopen(argv[1], "r")))
{
perror("Error opening file!\n");
exit(1);
}
while ((c = fgetc(fp)) != EOF)
{
char regdtest[1];
char regltest[1];
regdtest[0] = (char)c;
reti_d = regexec(&digit, regdtest, 0, NULL, 0);
reti_l = regexec(&letter, regltest, 0, NULL, 0);
switch (c)
{
case '(': printf("lperen "); break;
case ')': printf("rparen "); break;
case '+': printf("plus "); break;
case '-': printf("minus "); break;
case '*': printf("times "); break;
}
if (c == ':')
{
if (c++ == '=')
printf("assign ");
else printf("\n%c cannot follow : (colon).\n", (char)c);
}
if (c == '/')
{
if (++c == '/' || ++c == '*')
{
while (c != '\n' || (c == '*' && ++c == '/'))
c++;
}
else printf("div ");
}
if (c == '.')
{
regdtest[0] = (char)++c;
reti_d = regexec(&digit, regdtest, 0, NULL, 0);
if (!reti_d)
{
printf("number ");
}
else printf("\n. (dot) should be followed by digits.\n");
}
if (!reti_d)
{
while (!reti_d)
{
regdtest[0] = (char)c;
reti_d = regexec(&digit, regdtest, 0, NULL, 0);
}
printf("number ");
}
if (!reti_l)
{
int i, j = 0;
char read[5], write[6];
read[i] = write[j] = (char)c;
while(!reti_d || !reti_l)
{
c++;
i++;
j++;
if (i >= 5)
i = 0;
if (j >= 6)
j = 0;
read[i] = write[j] = (char)c;
if (strcmp(read, "read") == 0)
printf("read ");
else printf("id ");
if (strcmp(write, "write") == 0)
printf("write ");
else printf("id ");
regdtest[0] = (char)c;
regltest[0] = (char)c;
reti_d = regexec(&digit, regdtest, 0, NULL, 0);
reti_l = regexec(&letter, regltest, 0, NULL, 0);
}
printf("letter ");
}
}
fclose(fp);
return 0;
}
答案 0 :(得分:1)
您需要再次致电fgetc()
来获得第二个角色。
答案 1 :(得分:0)
正如@pmg建议的那样,将++ c替换为fgetc()
。
以下是经过重新设计的if (c == '/')
部分。各种语法错误路径未完全编码,下面可能需要额外的cod来处理意外路径。请记住,任何fgetc()
都可能导致EOF条件。在ungetc()
撤消调用之间最多可执行1 fgetc()
。
if (c == '/') {
c = fgetc(fp);
if (c == '/') {
c = fgetc(fp);
if (c == '*') {
while (c = fgetc(fp) != '\n' && c != EOF) {
if (c == '*') {
c = fgetc(fp);
if (c == '/') break;
ungetc(c, fp);
}
}
printf("div ");
}
else TBD();
} else ...