1）case '@': ClearStack(S); break; 清空栈后，想再不初始化就再使用？top指针其实编程空指针了。2）什么情况下用变量的使用问题，typedef struct{。。。}SqStack,*S;中的*S，InitStack(S);把S当做全局指针？基础要加强。3）代码修正如下，可以正确运行。include<stdlib.h> include<stdio.h> ...
您prime()函数体}后没加；include <iostream> using namespace std;int prime(int n){ int i;for(n=2;n<2000;n++){ for(i=2;i<=2000;i++){if(n%i==0) break;} if(n/i==1) return n;} };int main(){ int m;int a=prime(int n);int b=prime(int n);for(m=2;m<=...
case '4':pre_hungry();break;case '5':quit();break;default:goto x;} system("pause");} void take(){ cout<<"申请左边筷子"<<endl;if(stickstate[i] == nused){ stickstate[i] == used;cout<<"拿起左边筷子"<<endl;} cout<<"申请右边筷子"<<endl;if(stickstate[i-1] == ...
没怎么细看楼主的程序，不过楼主的错误显而易见，error : expected primary-expression before ' || ' token error : expected ' : ' before ' printf '已经说明错误出在||附近，printf之前 if((num%2)!=0)||(num<=4)少了右括号。其实num%2和num<=4也不用再加括号，if的括号就够了。改...
增加头文件 include <string.h> 错误行改为：strcpy(eq[i],"'c','d','b','=','a','|','e'");
要么去掉分号，改成如下：cout<<"A 1-character name? Hmm,have you read Kafka?:"<<"hello,"<< user_name <<endl;要么改成如下：cout<<"A 1-character name? Hmm,have you read Kafka?:" ;cout <<"hello,"<< user_name <<endl;
int main() { string staffName; int staffProf; while (staffName != "done") { cout << "Please enter a staff name(done to leave the program):" << endl; getline(cin, staffName, '\n'); if (staffName == "done") { break; ...
break;} if(sum3<(10*wa))///括号用中文状态输入了，改成英文状态输入 {sum3+=c;} else {m=sum3%(wa*10);///括号用中文状态输入了，改成英文状态输入 l=(sum3-m)/(wa*10);sum3=m+c;} if(sum2<(wa*10))sum2+=b;else { m=sum2%(wa*10);k=(sum2-m)/(wa*10);sum2...
expected primary expression before '(' token. is notorious for not necessarily occurring on that line, but being a knock-on effect of things like imbalanced brackets long before (or occasionally after) the line that gets flagged. I can only suggest that you break your source file down into ...
#include <iostream>usingnamespacestd;structpet{ string name;intage;boolisNeutered; };voidprintPets(pet* Pets,inti) { cout << Pets[i].name <<" "; cout << Pets[i].age <<" ";switch(Pets[i].isNeutered) {casetrue: cout <<"Yes\n";break;casefalse: cout <<"No\n";break; } }int...