我们随时为您提供帮助。
所以让我首先尝试编译您的代码。这是我的 C++ 编译器的结果：
所以，我们现在已经可以看到几件事情了。 VLA（可变长度数组）int array[n]是无效的 C++ 代码。 VLA 不是 C++ 语言的一部分，不能使用。此外，编译器显示了未初始化的 varaibale `max' 的问题。如果 may 没有值，那么你的程序的行为是未定义的。
强烈推荐。请始终打开所有编译器警告。示例：-Werror -Wall -Wpedantic。并修复所有警告。
好的，让我们尝试摆脱这些问题。首先是 VLA。 VLA 始终可以替换为std::vector。这是一个动态数组，可以根据需要增长。但后来我经常听到学生说他们没有了解vector。这基本上意味着老师应该被解雇，但也许首先教学生使用原始指针用于拥有的内存和new，这在C++中基本上是禁止的，这可能是一个战略决定。但是，好的，让我们使用这种方法。最后不要忘记删除分配的内存。
并且必须初始化变量“max”。我们将在此处使用可能的最小值，以便我们可以确保其他所有内容都将大于此初始值。
在标题<limits> 中，我们可以找到我们需要的所有内容。请看here。
结果是这个已经在运行的程序：
#include <iostream>
#include <limits>
using namespace std;
int main() {
int max=std::numeric_limits<int>::min(), n, check; //declaring variables
check = 1;
while (check == 1) { //programm executes repitedly
cout << "Enter array element count: ";
cin >> n; //user inputs number of elements in array
if (n <= 0) //check that number should be bigger then 0
cout << "Please enter number greater then 0";
int *array = new int[n]; //declaring an array
for (int i = 0; i < n; i++) { //entering numbers to array
cout << "Enter number: ";
cin >> array[i];
if (max < array[i]) { //assigning the biggest number of array as max
max = array[i];
}
}
for (int i = 0; i < n; i++) { //assigning value to max==1
if (array[i] == max) {
array[i] = 1;
}
else { // to other value assigning 0
array[i] = 0;
}
cout << array[i]; //showing the results
}
cout << "\n1 - continue, 0 - end\n";
cin >> check;
delete[] array;
}
}
还有很多事情需要检查。特别是对 unser 输入的验证很重要。更强大的版本是：
#include <iostream>
#include <limits>
int main() {
// We will rund the program until somebody sets this variable to false
bool runProgram = true;
while (runProgram) {
// Instruction to use, to enter the number of elements to check
std::cout << "\nEnter array element count: ";
// Get the number of elements and check, if the we got a valid number
unsigned int arrayElementCount = 0;
if ((std::cin >> arrayElementCount) and (arrayElementCount > 0u)) {
// OK we got a valid number. Now allocate memory for our array.
int* array = new int[arrayElementCount];
// We want to find the max value for all values. Therefore we will initialie the initial value with min
int max = std::numeric_limits<int>::min();
// In a loop, get user input, store in array and check for max
for (unsigned int i = 0; i < arrayElementCount; ++i) {
// We want to check, if the user enters a valid number or some nonesend like "abc"
// So, we will read as long the number is not valid
bool numberIsValid = false;
while (not numberIsValid) {
// Instruction to user to enter number i
std::cout << "Enter number '" << i + 1 << "' :\t";
// Try to read a valid number
if (std::cin >> array[i]) {
// Oke, noe we have a valid number. We do not need to read again
numberIsValid = true;
// CHeck for a new maximum
if (max < array[i]) {
max = array[i];
}
}
else {
// We got a bad number. std::cin is in error state and the input buffer contains garbage
// Show error message
std::cerr << "\n*** Error: Invalid input. Please try again\n\n";
std::cin.clear(); // Reset error flages of std::cin and remove still existing garbage from the input buffer
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
}
}
// We want to show the result
std::cout << "\n\nResult:\n\n";
for (unsigned int i = 0; i < arrayElementCount; i++) {
// For a max value we will write a 1 to our array
if (array[i] == max) {
array[i] = 1;
}
else {
// To other value assign a 0
array[i] = 0;
}
// And output restult
std::cout << array[i] << ' ';
}
// Ask user, if he wants to continue
std::cout << "\n\n\n1 - continue program, 0 - end program\n";
// Get decision of user
int check = 0;
// In case of error or other input than 1, end program
if (not (std::cin >> check) or (check != 1))
runProgram = false;
// Release previously allocated memory
delete[] array;
}
else {
// Error while getting the number of elements for the array
std::cerr << "\n*** Error: Problem with the input of the element count\n\n";
std::cin.clear();// Reset error flags of std::cin and remove still existing garbage from the input buffer
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
}
}
如有问题，请在评论中提问