爲什麼這個循環會被忽略？

Question

我不知道我在做什麼錯，但for循環沒有初始化
代碼只是立即顯示printfs。這其中沒有任何值，因爲for循環沒有激活

#define _CRT_SECURE_NO_WARNINGS 
#include <stdio.h> 
#include <stdlib.h> 
#define PAUSE system("Pause") 

main() { 
    // INITALIZE VARIABLES 
    int number = 0; 
    int i = 0; 
    int odd = 0; 
    int even = 0; 
    int totalNum = 0; 
    int tempNum = 0; 
    int count; 

    printf("Enter a number between 2 and 25\n"); 
    scanf("%i", &number); 

    do{ 
     if (number < 2 || number > 25) 
      printf("That was an invalid number please try again\n"); 
      scanf("%i", &number); 
    } while (number < 2 || number > 25); 

    printf("Enter how many numbers you want to input\n"); 
    scanf("%i", &count); 

    for (i = 1; i == count; ++i){ 
     printf("input numbers\n"); 
     scanf("%i", &tempNum); 
     if (tempNum % 2 == 0) 
      even++; 
     else 
      odd++; 
     totalNum = totalNum + tempNum; 
    } // END FOR LOOP 

    // DISPLAY OUTPUT 
    printf("You entered %i numbers\n", count); 
    printf("The sum of the %i numbers is %i\n", count, totalNum); 
    printf("The average of the %i numbers is %i\n", count, totalNum/count); 
    printf("You entered %i odd numbers and %i even numbers\n", odd, even); 

    PAUSE; 
} // END MAIN 

Answer 1

i = 1，所以i == count;給false因此循環被忽略。

Answer 2

當您初始化i至1時，您的循環最多隻能執行一次，當時爲count == 1。

如果你輸入計數1，

printf("Enter how many numbers you want to input\n"); 
    scanf("%i", &count); 

循環將剛好運行一次，直到i遞增到2

你可能想：

for (i = 1; i <= count; ++i){ 

Answer 3

for循環中C的作品如下：

for (variable initialization; condition; variable update) { 
    /* Do something... */ 
} 

該循環將執行長達條件爲爲真。所以，當你這樣做：

for (i = 1; i == count; ++i) 

的循環將執行，只要我==算是真實的。所以，除非count保持當這條線被執行時，循環將永遠不會運行。

正如其他人指出的那樣，你可能希望這樣的：

for (i = 1; i <= count; ++i) 

所以，你的循環將用於我所有值運行，直到它達到計數。

作爲一個方面說明我應該指出的是，通常的方式在C循環寫的是這樣的：

for (i = 0; i < count; i++) 

我們先從I = 0因爲C數組從零開始，所以數組中的第N個元素有索引n-1

Answer 4

你太親近了。除了修復您的循環測試條款for (i = 1; i <= count; i++)之外，我還建議您使用" %d"代替格式說明符。您的do循環只需要一個while循環以避免每次打印您的invalid number消息。

此外，儘管沒有錯誤，但C的標準編碼風格避免了caMelCase變量，而傾向於所有的小寫字母。見例如NASA - C Style Guide, 1994。

通過這些更改（並將您的奇數/偶數檢查更改爲簡單的&），您可以按如下方式編寫代碼。

#define _CRT_SECURE_NO_WARNINGS 
#include <stdio.h> 
#include <stdlib.h> 
// #define PAUSE system("Pause") 

int main (void) 
{ 
    int number, i, odd, even, totalnum, tempnum, count; 
    number = i = odd = even = totalnum = tempnum = count = 0; 

    printf ("enter a number between 2 and 25: "); 
    scanf (" %d", &number); 

    while (number < 2 || number > 25) { 
     printf ("invalid number, again (between 2 and 25): "); 
     scanf (" %d", &number); 
    } 

    printf ("numbers to input: "); 
    scanf (" %d", &count); 

    for (i = 1; i <= count; i++) { 
     printf ("input number %2d: ", i); 
     scanf (" %d", &tempnum); 
     if ((tempnum & 1) == 0) 
      even++; 
     else 
      odd++; 
     totalnum = totalnum + tempnum; 
    } 

    printf ("You entered %d numbers\n", count); 
    printf ("The sum of the %d numbers is %d\n", 
      count, totalnum); 
    printf ("The average of the %d numbers is %d\n", 
      count, totalnum/count); 
    printf ("You entered %d odd numbers and %d even numbers\n", 
      odd, even); 

//  PAUSE; 
    return 0; 
} 

注：main是類型int（例如int main (int argc, char **argv)或簡單地int main (void)，以指示沒有采取參數）。由於它是int類型，因此它會爲return一個值。儘管歷史實現可能允許void main，而便攜式代碼已不再適用。

示例使用/輸出

$ /bin/forskipped 
enter a number between 2 and 25: 4 
numbers to input: 4 
input number 1: 1 
input number 2: 2 
input number 3: 3 
input number 4: 4 
You entered 4 numbers 
The sum of the 4 numbers is 10 
The average of the 4 numbers is 2 
You entered 2 odd numbers and 2 even numbers 

看一下它，並讓我知道如果您有任何問題。

Answer 5

do{ 
    if (number < 2 || number > 25) 
     printf("That was an invalid number please try again\n"); 
     scanf("%i", &number); 
} while (number < 2 || number > 25); 

應該...

do{ 
    if (number < 2 || number > 25){ 
     printf("That was an invalid number please try again\n"); 
     scanf("%i", &number); 
    } 
} while (number < 2 || number > 25); 

否則它會詢問總是另一個號碼