strcpy及其第二个参数出错

Question

当我尝试编译这个程序时,我得到关于strcpy的第二个参数的错误(包含在代码下面).我老老实实地难以理解如何解决这个问题.我很抱歉,如果我的代码效率不高或看起来不好看; 我只是一名CS学生.

#include "stdafx.h"
#include <iostream>
#include <ctime>
using namespace std;

int main(){   


 int r = 0;
 char *article[]={"the", "a", "one", "some", "any"};
 char *noun[]={"boy","girl","dog","town","car"};
 char *verb[]={"drove","jumped","ran","walked","skipped"};
    char *preposition[]={"to","from","over","under","on"};
    char sentence [80];

 srand(time(NULL));
 for(int i=0;i<=20;i++){

    r = (rand()%5);
 strcpy(sentence,*article[r]);
 strcat(sentence," ");
    r = (rand()%5);
 strcat(sentence,*noun[r]);
 strcat(sentence," ");
 r = (rand()%5);
 strcat(sentence,*verb[r]);
 strcat(sentence," ");
 r = (rand()%5);
 strcat(sentence,*preposition[r]);
 strcat(sentence," ");
 r = (rand()%5);
 strcat(sentence,*article[r]);
 strcat(sentence," ");
 r = (rand()%5);
 strcat(sentence,*noun[r]);
 strcat(sentence,".");
 }

 sentence[0]= toupper(sentence[0]);
 cout<<sentence <<endl;


 system("pause");
 return 0;}

1>Compiling...
1>assignment 8.cpp
1>e:\assignment 8\assignment 8\assignment 8.cpp(16) : warning C4244: 'argument' : conversion from 'time_t' to 'unsigned int', possible loss of data
1>e:\assignment 8\assignment 8\assignment 8.cpp(20) : error C2664: 'strcpy' : cannot convert parameter 2 from 'char' to 'const char *'
1>        Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
1>e:\assignment 8\assignment 8\assignment 8.cpp(23) : error C2664: 'strcat' : cannot convert parameter 2 from 'char' to 'const char *'
1>        Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
1>e:\assignment 8\assignment 8\assignment 8.cpp(26) : error C2664: 'strcat' : cannot convert parameter 2 from 'char' to 'const char *'
1>        Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
1>e:\assignment 8\assignment 8\assignment 8.cpp(29) : error C2664: 'strcat' : cannot convert parameter 2 from 'char' to 'const char *'
1>        Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
1>e:\assignment 8\assignment 8\assignment 8.cpp(32) : error C2664: 'strcat' : cannot convert parameter 2 from 'char' to 'const char *'
1>        Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
1>e:\assignment 8\assignment 8\assignment 8.cpp(35) : error C2664: 'strcat' : cannot convert parameter 2 from 'char' to 'const char *'
1>        Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast

Answer 1

危险.strcat()并且strcpy()是代码癌症的主要原因.使用它们会使您暴露于各种缓冲区溢出.使用strncat()/ strncpy(),或者(甚至更好)只使用std::string,因为你正在使用C++!

strcat()并strcpy()期望他们的论点是字符串.*article[r]是单一的char- article[r]是你想要的字符串.所以,放下主要的星号.