8.编写一个程序,以一个字符和任意文件名作为命令行参数。如果字符后面没有参数,该程序读取标
淮输入;否则,程序依次打开每个文件并报告每个文件中该字符出现的次数。文件名和字符本身也要一同报告。程序应包含错误检查,以确定参数数量是否正确和是否能打开文件。如果无法打开文件,程序应报告这一情况,然后继续处理下一个文件。
int main(int argc, char* argv[])
{system("chcp 65001");if (argc < 2){fprintf(stderr, "用法: %s <文件1> [文件2] [文件3] ...\n", argv[0]);}else{if (argc == 2){char buf[1024];printf("请输入字符串:\n");scanf_s("%s", buf, 1024);int count = 0;for (int i = 0;i < strlen(buf);i++){if(buf[i] == argv[1][0])count++;}printf("字符%c出现的次数为%d\n", argv[1][0], count);}else{for (int i = 2;i < argc;i++){FILE* fp = NULL;fopen_s(&fp, argv[i], "rb");if(fp == NULL)printf("文件%s打开失败\n", argv[i]);else{int count = 0;char buf[1024];while (fgets(buf, sizeof(buf), fp) != NULL){for (int i = 0;i < strlen(buf);i++){if (buf[i] == argv[1][0])count++;}}printf("字符%c再文件%s中出现的次数为%d\n", argv[1][0], argv[i], count);fclose(fp);}}}}return 0;
}
9.修改程序清单 13.3 中的程序,从1开始,根据加入列表的顺序为每个单词编号。当程序下次运行时,确保新的单词编号接着上次的编号开始。
#define MAX 41
int main(void)
{FILE *fp;char words[MAX];int lineNo = 0;fopen_s(&fp, "wordy", "r");if (fp){while (fgets(words, MAX, fp) != NULL){ lineNo++;}fclose(fp);}fopen_s(&fp, "wordy", "a+");if (fp == NULL)fprintf(stdout, "Can't open \"wordy\" file.\n");else{puts("Enter words to add to the file; press the #");puts("Key at the beginning of a line to terminate.");while ((fscanf_s(stdin, "%40s", words, 41) == 1) && (words[0] != '#'))fprintf(fp, "%d %s\n", ++lineNo, words);rewind(fp);while (fscanf_s(fp, "%d %s", &lineNo, words, 41) == 2)printf("%d %s\n", lineNo, words);puts("Done.");fclose(fp);}
}
10.编写一个程序打开一个文本文件,通过交互方式获得文件名。通过一个循环,提示用户输入一个文件位置。然后该程序打印从该位置开始到下一个换行符之前的内容。用户输入负数或非数值字符可以结束输入循环。
int main()
{system("chcp 65001");puts("请输入文件名:");char filename[256];FILE *fp;scanf_s("%s", filename, 256);fopen_s(&fp, filename, "r");if (fp == NULL){puts("无法打开文件!");}else{puts("请输入要打印的位置(负数或非数字退出):");int pos;while (scanf_s("%d", &pos) == 1 && pos >= 0){fseek(fp, pos, SEEK_SET);puts(fgets(filename, 256, fp));puts("请输入要打印的位置(负数或非数字退出):");}}fclose(fp);return 0;
}
11.编写一个程序,接受两个命令行参数。第1个参数是一个字符串,第2个参数是一个文件名。然后该程序查找该文件,打印文件中包含该字符串的所有行。因为该任务是面向行而不是面向字符的,所以要使用fgets()而不是getc()。使用标准C库函数strstr()(11.5.7节简要介绍过)在每一行中查找指定字符串。假设文件中的所有行都不超过255个字符。
int main(int argc, char* argv[])
{system("chcp 65001");if (argc < 2){fprintf(stderr, "用法: %s <字符串> <文件名>\n", argv[0]);}else{FILE *fp;fopen_s(&fp, argv[2], "rb");if (fp){ char buf[1024];while (fgets(buf, sizeof(buf), fp)){if (strstr(buf, argv[1])){printf("%s\n", buf);}}fclose(fp);}elsefprintf(stderr, "无法打开文件 %s\n", argv[2]);}return 0;
}
12.创建一个文本文件,内含20行,每行30个整数。这些整数都在0~9之间,用空格分开。该文件是用数字表示一张图片,0~9表示逐渐增加的灰度。编写一个程序,把文件中的内容读入一个20X30的int数组中。一种把这些数字转换为图片的粗略方法是:该程序使用数组中的值初始化一个20X31的字符数组,用值0对应空格字符,1对应点字符,以此类推。数字越大表示字得所占的空间越大。例如,用#表示9。每行的最后一个字符(第31个)是空字符,这样该数组包含了20个字符串。最后,程序显示最终的图片(即,打印所有的字符串),并将给果存储在文本文件中。例如,下面是开始的数据:
0 0 9 0 0 0 0 0 0 0 0 0 5 8 9 9 8 5 2 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 9 0 0 0 0 0 0 0 5 8 9 9 8 5 5 2 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 5 8 1 9 8 5 4 5 2 0 0 0 0 0 0 0 0 0
0 0 0 0 9 0 0 0 0 0 0 0 5 8 9 9 8 5 0 4 5 2 0 0 0 0 0 0 0 0
0 0 9 0 0 0 0 0 0 0 0 0 5 8 9 9 8 5 0 0 4 5 2 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 5 8 1 9 8 5 0 0 0 4 5 2 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 5 8 1 9 8 5 0 0 0 0 4 5 2 0 0 0 0 0
5 5 5 5 5 5 5 5 5 5 5 5 5 8 9 9 8 5 5 5 5 5 5 5 5 5 5 5 5 5
8 8 8 8 8 8 8 8 8 8 8 8 5 8 9 9 8 5 8 8 8 8 8 8 8 8 8 8 8 8
9 9 9 9 0 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 3 9 9 9 9 9 9 9 9
8 8 8 8 8 8 8 8 8 8 8 8 5 8 9 9 8 5 8 8 8 8 8 8 8 8 8 8 8 8
5 5 5 5 5 5 5 5 5 5 5 5 5 8 9 9 8 5 5 5 5 5 5 5 5 5 5 5 5 5
0 0 0 0 0 0 0 0 0 0 0 0 5 8 9 9 8 5 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 5 8 9 9 8 5 0 0 0 0 6 6 0 0 0 0 0 0
0 0 0 0 2 2 0 0 0 0 0 0 5 8 9 9 8 5 0 0 5 6 0 0 6 5 0 0 0 0
0 0 0 0 3 3 0 0 0 0 0 0 5 8 9 9 8 5 0 5 6 1 1 1 1 6 5 0 0 0
0 0 0 0 4 4 0 0 0 0 0 0 5 8 9 9 8 5 0 0 5 6 0 0 6 5 0 0 0 0
0 0 0 0 5 5 0 0 0 0 0 0 5 8 9 9 8 5 0 0 0 0 6 6 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 5 8 9 9 8 5 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 5 8 9 9 8 5 0 0 0 0 0 0 0 0 0 0 0 0
根据以上描述选择特定的输出字符,最终输入如下:
# *%##%*'
# *%##%**'
*%.#%*~*'
# *%##%* ~*'
# *%##%* ~*'
*%.#%* ~*'
*%.#%* ~*'
*************%##%*************
%%%%%%%%%%%%*%##%*%%%%%%%%%%%%
#### ################:########
%%%%%%%%%%%%*%##%*%%%%%%%%%%%%
*************%##%*************
*%##%*
*%##%* ==
'' *%##%* *= =*
:: *%##%* *=....=*
~~ *%##%* *= =*
** *%##%* ==
*%##%*
*%##%*
int main(int argc, char* argv[])
{const char imgChar[] = { ' ', '.', '\'', ':', '~', '*', '=', '-', '%', '#'};char arr[20][31] = { 0 };FILE* fp;fopen_s(&fp, "img.txt", "rb");if(fp == NULL)printf("open file error!");else{char ch;int row = 0;int col = 0;while ((ch = getc(fp)) != EOF){if (ch >= '0' && ch <= '9'){arr[row][col++] = ch;}else if (ch == '\n'){row++;col = 0;}}fclose(fp);fopen_s(&fp, "imgout.txt", "wb+");for (int i = 0; i < 20; i++){for (int j = 0; j < 30; j++){arr[i][j] = imgChar[arr[i][j] - '0'];printf("%c", arr[i][j]);if (fp){fputc(arr[i][j], fp);}}printf("\n");if(fp){fputc('\n', fp);}}if(fp)fclose(fp);}return 0;
}
13.用变长数组(VLA)代替标准数组,完成编程练习12。
int main(int argc, char* argv[])
{const char imgChar[] = { ' ', '.', '\'', ':', '~', '*', '=', '-', '%', '#'};char **arr;FILE* fp;fopen_s(&fp, "img.txt", "rb");if(fp == NULL)printf("open file error!");else{int total_rows = 0;int max_cols = 0;char buf[1024];while (fgets(buf, 1024, fp)){total_rows++;int cols = strlen(buf) / 2;if (cols > max_cols)max_cols = cols;}rewind(fp);arr = (char**)malloc(total_rows * sizeof(char*));for (int i = 0; i < total_rows; i++){arr[i] = (char*)malloc(max_cols * sizeof(char));}char ch;int row = 0;int col = 0;while ((ch = getc(fp)) != EOF){if (ch >= '0' && ch <= '9'){arr[row][col++] = ch;}else if (ch == '\n'){row++;col = 0;}}fclose(fp);fopen_s(&fp, "imgout.txt", "wb+");for (int i = 0; i < total_rows; i++){for (int j = 0; j < max_cols; j++){arr[i][j] = imgChar[arr[i][j] - '0'];printf("%c", arr[i][j]);if (fp){fputc(arr[i][j], fp);}}printf("\n");if(fp){fputc('\n', fp);}}for (int i = 0; i < total_rows; i++){free(arr[i]);}free(arr);if(fp)fclose(fp);}return 0;
}
14.数字图像,尤其是从字宙飞船发回的数字图像,可能会包含一些失真。为编程练习12添加消除失真的函数。该函数把每个值与它上下左右相邻的值作比较,如果该值与其周围相邻值的差都大于1,则用所有相邻值的平均值(四舍五入为整数)代替该值。注意,与边界上的点相邻的点少于个,所以做特殊处理。
void remove_distortion(char** array, int rows, int cols) {// 创建临时数组保存原始值int** temp = (int**)malloc(rows * sizeof(int*));for (int i = 0; i < rows; i++) {temp[i] = (int*)malloc(cols * sizeof(int));for (int j = 0; j < cols; j++) {temp[i][j] = array[i][j];}}// 遍历所有点(包括边界)for (int i = 0; i < rows; i++) {for (int j = 0; j < cols; j++) {int current = temp[i][j];int neighbor_count = 0; // 实际相邻点数量int neighbor_sum = 0; // 相邻点值总和int all_diff_gt1 = 1; // 假设所有差值>1// 检查上邻居if (i > 0) {int up = temp[i - 1][j];neighbor_sum += up;neighbor_count++;if (abs(current - up) <= 1) all_diff_gt1 = 0;}// 检查下邻居if (i < rows - 1) {int down = temp[i + 1][j];neighbor_sum += down;neighbor_count++;if (abs(current - down) <= 1) all_diff_gt1 = 0;}// 检查左邻居if (j > 0) {int left = temp[i][j - 1];neighbor_sum += left;neighbor_count++;if (abs(current - left) <= 1) all_diff_gt1 = 0;}// 检查右邻居if (j < cols - 1) {int right = temp[i][j + 1];neighbor_sum += right;neighbor_count++;if (abs(current - right) <= 1) all_diff_gt1 = 0;}// 如果存在相邻点且所有差值>1,则替换当前值if (neighbor_count > 0 && all_diff_gt1) {// 计算平均值(四舍五入)double avg = (double)neighbor_sum / neighbor_count;array[i][j] = (int)round(avg);}}}// 释放临时数组内存for (int i = 0; i < rows; i++) {free(temp[i]);}free(temp);
}
int main(int argc, char* argv[])
{const char imgChar[] = { ' ', '.', '\'', ':', '~', '*', '=', '-', '%', '#'};char **arr;FILE* fp;fopen_s(&fp, "img.txt", "rb");if(fp == NULL)printf("open file error!");else{int total_rows = 0;int max_cols = 0;char buf[1024];while (fgets(buf, 1024, fp)){total_rows++;int cols = strlen(buf) / 2;if (cols > max_cols)max_cols = cols;}rewind(fp);arr = (char**)malloc(total_rows * sizeof(char*));for (int i = 0; i < total_rows; i++){arr[i] = (char*)malloc(max_cols * sizeof(char));}char ch;int row = 0;int col = 0;while ((ch = getc(fp)) != EOF){if (ch >= '0' && ch <= '9'){arr[row][col++] = ch;}else if (ch == '\n'){row++;col = 0;}}fclose(fp);remove_distortion(arr, total_rows, max_cols);fopen_s(&fp, "imgout.txt", "wb+");for (int i = 0; i < total_rows; i++){for (int j = 0; j < max_cols; j++){arr[i][j] = imgChar[arr[i][j] - '0'];printf("%c", arr[i][j]);if (fp){fputc(arr[i][j], fp);}}printf("\n");if(fp){fputc('\n', fp);}}if(fp)fclose(fp);}return 0;
}