如何從數組中獲取字符並將其存儲在變量中？

Question

#include <stdio.h> 
#include <stdlib.h> 

int main() 
{ 

    int agecalc; 
    int nextyr; 
    int birthday; 
    int currentday; 
    int agecalcu; 
    int randomnumbers; 

    birthday = 1987; 
    currentday = 2016; 
    nextyr = currentday + 1; 
    agecalc = currentday - birthday; 
    randomnumbers = 7890; 

    char My[] = "Sayan Banerjee"; 

    printf("%s 's birthday is %.3d \n", My , agecalc); 

    agecalcu = agecalc + 1; 

/* alternatively I can use one int and change the variable stored in agecalc by stating that 
agecalc = currentday - birthday +1; But I think that is just messy and disturbs a lot of code. declaring a new int is 
better and less intrusive. this way when I may need to use agecalc again, i dont have to worry about the value stored in that 
variable. */ 

    printf("%.5s will turn %.3d on %d.\n \a", My , agecalcu , nextyr); 

    printf("The username for %s is %.6d \n", My , randomnumbers); 
    // the thing here is that I cannot shorten a value stored in a value unless its in an array 
    // but I can express it in other ways like 1 to 01. 
    /* The thing is that I cannot store two %s in one argument/code. this freezes the code. better way would be to 
    create another variable and then try to edit that.*/ 

    //this is an experiment to see whether I can take characters from an array and store it in a variable 
    int user; 


    My [0,1,2,3,4,5] = user; 

    printf("%s is also %s", My , user); 


    return 0; 
} 

我的主要問題是在最後一行。我是一名編程新手，剛剛開始學習。我只是在玩耍，注意到如果我在同一個參數中輸入兩個%s，程序就會崩潰。所以，我在考慮是否可以將數組My中的某些字符存儲在變量中，然後將其打印出來？

可能嗎？或者我只是以錯誤的方式看待這個問題？ 對不起，這個雜亂的帖子。 感謝您的幫助。

Answer 1

My [0,1,2,3,4,5] = user;

這一行相當於說

My[5] = user;

這是因爲comma操作

//this is an experiment to see whether I can take characters from an array and store it in a variable 

你可以的。但是你現在正在做的是將一個未分配的值分配給My陣列的5th元素。 int val = My[5]是完全正確的，而不是其他方式。

飛機墜毀發生，因爲你的代碼試圖解釋一個整數作爲null terminated string因爲%s格式說明你目前的狀態給

printf("%s is also %d", My , user); // %d for integers 

你的代碼是在Undefined Behavior土地。

若要將現有數組中的某些字符複製到目標數組中，您將需要多個變量。一個變量只會保存一個字符。

char user[6]; 
user[5] = '\0'; // fill the last character as null to be a valid C string 
user[0] = My[0]; 
user[1] = My[1]; 
user[<any other index you want>] = My[<The index you want>]; 

您可以通過簡單地使用從string.h其中有一堆字符串處理函數的實用程序保存自己的精力，時間和精力。可能你目前的問題需要的是strncpy

作爲一個正交的建議，編譯你的代碼時使用編譯器支持的最高警告級別。編譯器會在編譯時提出這些問題。

對於如gcc -Wall filename.c

Answer 2

行之後不會做你認爲：

這樣的：

My [0,1,2,3,4,5] = user; 

其實是同樣的事情，這樣的：

My [5] = user; 

閱讀有關comma operator。

無論如何，您不能將整個My數組存儲在單個int變量中。

---

這裏使用的是%s打印user這不是一個char*而是int

printf("%s is also %s", My , user); 

這將導致不確定的行爲，很有可能是在現代系統崩潰。

---

至於你的問題實在是有點不清楚，你也許想要這個：

#include <string.h> 
... 
char user[20]; 
strcpy(user, My); 

printf("%s is also %s", My , user); 

，它將打印：

Sayan Banerjee is also Sayan Banerjee