c - 去掉C中的所有注释

Question

现在此代码不会删除内联注释，我该如何更改它以便它也删除内联注释？

FILE *output;
output = fopen("preprocess_output.c", "w");

while (fgets(line, LINE_LENGTH, file) != NULL)
{
    for (int i = 0; i < strlen(line); i++)
    {
        if (line[i]  ==  '/' && line[i + 1]  ==  '/')
        {
            comment_lines++;
        }
        else
        {
            fprintf(output, line);
        }
        if (line[i] != '\n' && line[i] != '\t')
        {
            non_blank++;
            break;
        }   
    }
}

Answer 1

这是一个小程序，几乎在所有情况下都会删除 C 注释。

/* strip C comments by chqrlie */

#include <errno.h>
#include <stdio.h>
#include <string.h>

/* read the next byte from the C source file, handing escaped newlines */
int getcpp(FILE *fp, int *lineno_p) {
    int ch;
    while ((ch = getc(fp)) == '\\') {
        if ((ch = getc(fp)) != '\n') {
            ungetc(ch, fp);
            return '\\';
        }
        *lineno_p += 1;
    }
    if (ch == '\n')
        *lineno_p += 1;
    return ch;
}

int main(int argc, char *argv[]) {
    FILE *fp = stdin, *ft = stdout;
    const char *filename = "<stdin>";
    int ch, lineno;

    if (argc > 1) {
        if ((fp = fopen(filename = argv[1], "r")) == NULL) {
            fprintf(stderr, "Cannot open input file %s: %s\n",
                    filename, strerror(errno));
            return 1;
        }
    }
    if (argc > 2) {
        if ((ft = fopen(argv[2], "w")) == NULL) {
            fprintf(stderr, "Cannot open output file %s: %s\n",
                    argv[2], strerror(errno));
            return 1;
        }
    }
    lineno = 1;
    while ((ch = getcpp(fp, &lineno)) != EOF) {
        int startline = lineno;
        if (ch == '/') {
            if ((ch = getcpp(fp, &lineno)) == '/') {
                /* single-line comment */
                while ((ch = getcpp(fp, &lineno)) != EOF && ch != '\n')
                    continue;
                if (ch == EOF) {
                    fprintf(stderr, "%s:%d: unterminated single line comment\n",
                            filename, startline);
                    break;
                }
                putc('\n', ft);  /* replace comment with newline */
                continue;
            }
            if (ch == '*') {
                /* multi-line comment */
                int lastc = 0;
                while ((ch = getcpp(fp, &lineno)) != EOF) {
                    if (ch == '/' && lastc == '*') {
                        break;
                    }
                    lastc = ch;
                }
                if (ch == EOF) {
                    fprintf(stderr, "%s:%d: unterminated comment\n",
                            filename, startline);
                    break;
                }
                putc(' ', ft);  /* replace comment with single space */
                continue;
            }
            putc('/', ft);
            /* keep parsing to handle n/"a//"[i] */
        }
        if (ch == '\'' || ch == '"') {
            int sep = ch;
            const char *const_type = (ch == '"') ? "string" : "character";

            putc(sep, ft);
            while ((ch = getcpp(fp, &lineno)) != EOF) {
                putc(ch, ft);
                if (ch == sep)
                    break;;
                if (ch == '\\') {
                    if ((ch = getcpp(fp, &lineno)) == EOF)
                        break;
                    putc(ch, ft);
                }
                if (ch == '\n') {
                    fprintf(stderr, "%s:%d: unescaped newline in %s constant\n",
                            filename, lineno - 1, const_type);
                    /* This is a syntax error but keep going as if constant was terminated */
                    break;
                }
            }
            if (ch == EOF) {
                fprintf(stderr, "%s:%d: unterminated %s constant\n",
                        filename, startline, const_type);
                break;
            }
            continue;
        }
        putc(ch, ft);
    }
    if (fp != stdin)
        fclose(fp);
    if (ft != stdout)
        fclose(ft);
    return 0;
}

既然您免费获得了完整的答案，请尝试了解上面的代码如何处理字符串和转义换行符。还有一些不支持的极端情况，你能找到它们吗？

• 其中一个极端情况是代码不解析三字母，这是一种过时的功能，可用于隐藏 \ 字符。