sublime text C++編譯會導致seg fault

Question

我最近從windows上的dev C++切換到osx dev環境，並試圖使用崇高的文本3.但是，當我運行我的程序時，出現了分段錯誤。以下是錯誤消息：

/bin/bash: line 1: 20506 Segmentation fault: 11 "/Users/jimi/Documents/University/lab0603" 
[Finished in 1.2s with exit code 139] 
[shell_cmd: g++ "/Users/jimi/Documents/University/lab0603.cpp" -o "/Users/jimi/Documents/University/lab0603" && "/Users/jimi/Documents/University/lab0603"] 
[dir: /Users/jimi/Documents/University] 
[path: /usr/bin:/bin:/usr/sbin:/sbin] 

下面是完整的代碼：

#include <iostream> 
#include <cmath> 

using namespace std; 


string flightDirections[16] = {"ENE", "NE", "NNE", "N", "NNW", "NW", "WNW", "W", "WSW", "SW", "SWS", "S", "SSE", "SE", "ESE", "E"}; 
double cruisingSpeed = 0; 
double windSpeed = 0; 
int windDirection = 0; 
double flightDistance = 0; 

string numberToDirection(int direction) { 

    return (direction) ? "The wind is blowing from the East." : " The wind is blowing from the West."; 

    //this way we only have one return statement :) 

} 

void getInput() { 
    cout << "Hello! \n" 
     << "Thank you for choosing to use this really great program!! \n" 
     << "This program will compute the necessary heading adjustment for your flight," 
     << " and provide the estimated flight time. \n"; 
    cout << "Enter the aircraft cruising speed in still air (in km/h): "; 
    cin >> cruisingSpeed; 
    cout << " \n \t cruising speed = " << cruisingSpeed << "\n Enter the wind speed in km/h: "; 
    cin >> windSpeed; 
    cout << " \n \t wind speed = " << windSpeed << "\n Enter 1 if the wind is blowing from the West and -1 if wind is blowing from the East:"; 
    cin >> windDirection; 
    cout << "\n\t" << numberToDirection(windDirection) << "\n Enter the distance between the originating and destination cities, in km:"; 
    cin >> flightDistance; 
    cout << "\n\t flight distance = " << flightDistance << "\n Enter the compass direction of the destination city, relative to the originating cities, using the following values:"; 
    for (int i = 0; i < sizeof(flightDirections); i++) { 
     cout << i + 1; 
     cout << flightDirections[i]; 
    } 
    cin.ignore(); 


} 

int main() { 


    getInput(); 
    return 0; 

} 

這是怎麼回事？

Answer 1

您錯誤地使用了sizeof。 sizeof返回實體包含的字節數，而不是條目數（對於數組的情況）。因此，這是不正確的：

for (int i = 0; i < sizeof(flightDirections); i++) { 

你應該簡單地使用：

for (int i = 0; i < sizeof(flightDirections)/sizeof(flightDirections[0]); i++) { 

Answer 2

你的問題是你的for循環。

for (int i = 0; i < sizeof(flightDirections); i++) 

將運行關閉陣列的端部作爲sizeof(flightDirections)是磨碎機比所述陣列的大小。 sizeof(flightDirections)是sizeof(std::string) * number_of_elements_in_the_array。爲了得到你需要使用

sizeof(flightDirections)/sizeof(flightDirections[0]) 

或者更好的使用數組的正確尺寸ranged based for loop作爲

int i = 0; 
for (const auto & e : flightDirections) { 
    cout << ++i << e; 
} 

Answer 3

sizeof給人以字節爲單位的大小，數組的元素數量不限。當你打電話給sizeof(flightDirections)時，你可能會預期16，但是驚慌更大。因此，你的循環太過分了，超出了你的數組的索引，並且你得到了未定義的行爲 - 在這種情況下，出現了分段錯誤。

由於你已經硬編碼了數組的大小，你甚至可以硬編碼循環。或者，如果你想擁有更靈活的代碼，你可以通過將的sizeof（flightDirections）的reuslt由單一元素的大小，類似這樣的計算元素的數量：

for (int i = 0; i < sizeof(flightDirections)/sizeof(flightDirections[0]); i++) { 

（這招很常見在C語言中練習時，你必須將一個數組傳遞給一個函數：因爲數組會失去它的大小（當它傳遞給指針時會衰減），你必須顯式地傳遞它作爲第二個參數，這是正常的方式）

Answer 4

如果要使用sizeof（）來查找數組的長度，則需要將它除以數組中的內容。看到這個問題：

How do I find the length of an array?

你可以使用一個std :: vector的呢？這有一個size（）方法。