#include<iostream> #include<complex.h> #include <complex> #include <math.h> using namespace std; #define pi 3.14159265358979323846 int main() { file* stream; stream = fopen("4000-red.raw","rb"); unsigned char* image = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); fread(image,4000*4000,sizeof(unsigned char),stream); unsigned char* image2 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); unsigned char* image3 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned
这段 C++ 代码看起来是在进行文件读取和内存分配相关的操作。其中存在一些错误和不完整的部分: 1. `file* stream;` 应该为 `FILE* stream;` ,注意大小写。 2. 代码中对于 `image3` 的内存分配没有完成,缺少 `sizeof(unsigned char)` 的参数。 以下是修正后的代码示例: ```cpp #include <iostream> #include <complex.h> #include <complex> #include <math.h> #include <stdio.h> // 包含 fopen、fread 等函数所需的头文件 #define pi 3.14159265358979323846 int main() { FILE* stream; stream = fopen("4000-red.raw","rb"); if (stream == NULL) { std::cout << "Failed to open the file." << std::endl; return 1; } unsigned char* image = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); fread(image, 4000 * 4000, sizeof(unsigned char), stream); unsigned char* image2 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); unsigned char* image3 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); // 在这里添加后续对分配的内存进行处理的代码 fclose(stream); free(image); free(image2); free(image3); return 0; } ``` 请问您是在学习文件操作和内存分配相关的知识,还是这段代码是您实际项目中的一部分,需要进一步完善其功能? [2024-11-02 18:16:35 | AI写代码神器 | 413点数解答]
- [字节豆包] #include<iostream> #include<complex.h> #include <complex> #include <math.h> using namespace std; #define pi 3.14159265358979323846 int main() { file* stream; stream = fopen("4000-red.raw","rb"); unsigned char* image = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); fread(image,4000*4000,sizeof(unsigned char),stream); unsigned char* image2 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned char)); unsigned char* image3 = (unsigned char*)malloc(4000 * 4000 * sizeof(unsigned(413点数解答 | 2024-11-02 18:16:35)300
- [字节豆包] 采用c++语言,实现如下功能: 实现古典密码中的移位密码和仿射密码,具体实现如下接口: c++接口: <<<<<移位密码>>>>> bool is_valid_s(unsinged char k) { 判断k是否为合理的密钥 } int encrypt_s(unsigned char* p, unsigned char k) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_s(unsigned char* c, unsigned char k) { 密钥合法则返回1,且明文覆盖密文: 密钥不合法则返回0. } <<<<<仿射密码>>>>> bool in_valid_a(unsinged char a, unsigned char b) { 判断a,b是否为合理的密钥 } int encrypt_a(unsigned char* p, unsigned char a, unsigned char b) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_a(unsigned char* c, unsign(812点数解答 | 2024-12-18 16:02:36)330
- [字节豆包] int n = 3; int width1 = width * n; int height1 = height * n; unsigned char* image9 = (unsigned char*)malloc(sizeof(unsigned char) * width1*height1); for (int i = 0; i < height1; i++) { for (int j = 0; j < width1; j++) { int a = i /n; int b = j /n; image9[i * width1 + j] = image1[a * width + b ]; } } char result6[100] = "suofang.raw"; file* fp6 = fopen(result6, "wb"); fwrite(image9, sizeof(unsigned cha(348点数解答 | 2024-10-26 12:13:04)274
- [阿里通义] #include<stdio.h> #include<malloc.h>//动态存储分配函数头文件 #include<math.h>//包含数学函数的文件 #include<string.h>//一个和字符串处理相关的头文件 #include<process.h>//包含用于和宏指令的作用声明 #define error 0 //宏定义 #define ok 1 #define over -2 #define listinitsize 20 #define listincrement 5 #define namelen 8 #define majorlen 20 #define init_n 3 typedef int status; //自定义类型语句 status i 等价于 int i typedef char statusc; typedef struct{ statusc number[10]; //学号 statusc name[namelen+1]; //姓名 statusc gender; //性别 statusc telephone[12]; //电话 st(1142点数解答 | 2024-11-07 19:08:54)241
- [GPT] 完善代码#include<reg51.h> #include <intrins.h> code unsigned char tab[] = {0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x40}; unsigned char dspbuf[8] = {0,0,10,0,0,10,0,0}; unsigned char i,j,k; unsigned char dspbuf1[8] = {0,0,10,0,0,10,0,0}; code unsigned int yindiao[]={0,120,106,94,89,79,71,62,59, 56, 53, 50, 47, 44, 41, 38}; code unsigned int jiepai[]={0,523,587,659,698,784,880,988};(334点数解答 | 2024-11-25 18:12:57)229
- [GPT] 告诉我那里错误#include<reg51.h> #include <intrins.h> code unsigned char tab[] = {0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x40}; unsigned char dspbuf[8] = {0,0,10,0,0,10,0,0}; unsigned char i,j,k; unsigned char dspbuf1[8] = {0,0,10,0,0,10,0,0}; code unsigned int yindiao[]={0,120,106,94,89,79,71,62,59, 56, 53, 50, 47, 44, 41, 38}; code unsigned int jiepai[]={0,523,587,659,698,784,880,988(78点数解答 | 2024-11-25 18:13:58)229
- [阿里通义] %% 1. 参数区——只改这里就行 excelFile = '运动者1的跳远位置信息.xlsx'; % 你的Excel文件名 sheetName = 1; % 工作表序号或名字 dt = 1/30; % 帧间隔,按实际改! savePic = true; % 是否自动保存图片 %% 2. 读取Excel(0_X 0_Y ... 31_X 31_Y) raw = readmatrix(excelFile, 'Sheet', sheetName); % N×64 [N, ~] = size(raw); %% 3. 整理成 N×32×2 raw = raw.'; % 64×N pos = reshape(raw, 2, 32, []); % 2×32×N pos = permute(pos, [3 2 1]); % N×32×2 %% 4. 算加速度(损失首尾两帧) ac(774点数解答 | 2025-09-04 21:03:12)93
- [DeepSeek] #include <iostream> #define MAXSIZE 100 #define OK 1 #define ERROR 0 #define OVERFLOW -2 using namespace std; typedef struct {//符号栈 char* base; char* top; int stacksize; }SqStack1; int InitStack1(SqStack1& S) {//符号栈初始化 return OK; } int Push1(SqStack1& S, char e) {//符号栈入栈 return OK; } int Pop1(SqStack1& S) {//符号栈出栈 return OK; } char GetTop1(SqStack1 S) {//符号栈取栈顶元素 return ERROR; } typedef struct {//数字栈 double* base; double* top; int stacksize; }SqStack2; int InitStack2(SqStack2&(1360点数解答 | 2025-11-14 14:57:43)88
- [字节豆包] #include <iostream> #include <sstream> #include <string> #include <cctype> #include <vector> // 将字符串转换为小写 std::string toLowerCase(const std::string& str) { std::string result = str; for (char& c : result) { c = std::tolower(c); } return result; } int main() { std::string target; std::getline(std::cin, target); target = toLowerCase(target); std::string article; std::getline(std::cin, article); std::stringstream ss(article); std::vector<std::(1381点数解答 | 2025-12-15 17:32:28)31
- [字节豆包] 题目描述 输出 <math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>n</mi> <mo>!</mo> </math> 的不同因子的个数? 输入 一个整数<math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>n</mi> </math> 输出 输出<math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>n</mi> <mo>!</mo> </math> 的因子个数 样例输入 复制 3 样例输出 复制 4 提示<math xmlns="http://www.w3.org/1998/Math/MathML"> <mn>1</mn> <mo><=</mo> <mi>n</mi> <mo><=</mo> <msup> <mn>10</mn> <mrow data-mjx-texclass="ORD"> <mn>5</mn> </mrow> <(731点数解答 | 2026-01-12 12:15:34)82
- [阿里通义] 把这个代码修改正确#include<reg52.h> sbit key = p3^0; // 按键连接到p3.0 unsigned char code seg_code[] = {0x3f, 0x06, 0x5b, 0x4f, 0x66, 0x6d, 0x7d, 0x07, 0x7f, 0x6f}; // 数码管段码 unsigned char code dash_code = 0x40; // 横杠段码 unsigned int second =0; // 秒表计数 bit running = 0; // 秒表运行状态 void timer0_isr() interrupt 1 { static unsigned int count =0; th0 = (65536 - 1000) / 256; tl0 = (65536 - 1000) %256; count++; if (count >= 1000) { count = 0; if (running) { second++; } } } void delay(unsigned int ms) { unsigned int i(776点数解答 | 2024-10-23 22:51:51)271
- [讯飞星火] 快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(710点数解答 | 2024-11-14 10:44:25)331