C++ 如何捕捉只有换行符或空格字符缓冲区的边缘大小写

我有以下代码

int * counter(const char *filename, bool bytes_flag, bool words_flag, bool lines_flag){
    int * counts = malloc(sizeof(int) * 3);
    int line_count = 0;
    int word_count = 0;
    int byte_count = 0;
    size_t line_size = LINE_MAX;
    FILE * file_stream;
    char * line_buffer;

    printf("%d %d %d\n", line_count, word_count, byte_count);

    file_stream = fopen(filename, "r");

    if (file_stream == NULL) {
        printf("File count not be opened.\n");
    }
    else{
        line_buffer = (char *)malloc(line_size);
        while (getline(&line_buffer, &line_size, file_stream) != EOF){

            if (lines_flag){line_count++;}

            for ( int i = -1; line_buffer[i] != '\n'; i++){
                byte_count++;
                if (isspace(line_buffer[i]) && !isspace(line_buffer[i-1])){
                    word_count++;
                }
            }
            //byte_count++;//Count last \n character
        }
        fclose(file_stream);
        
        printf("\n");
        printf("%d\n", line_count);
        printf("%d\n", word_count);
        printf("%d\n", byte_count);
        printf("\n");

        counts[0] = line_count;
        counts[1] = word_count;
        counts[2] = byte_count;
    }
    return counts;
}

我正在输入以下文件

The Project Gutenberg eBook of The Art of War
    
This ebook is for the use of anyone anywhere in the United States and

我的输出几乎就是我想要的，但字数1太高了.发生这种情况是因为文本文件第2行上的空行.

./ccwc.o -l small.txt
0 0 0

3
24
127

我如何才能提高？

if (isspace(line_buffer[i]) && !isspace(line_buffer[i-1])){
    word_count++;
}

才能抓住这个边缘 case ？