为什么我的代码返回一个不正确的值？

Question

我对C非常陌生，我想弄清楚为什么我的代码返回的值不正确。

int main() 
{ 


    printf("Welcome to my number generator! \n"); 
    printf("What is the first number in the range? \n"); 
    int rng1 = scanf("%d", &rng1); 
    printf("What is the second number in the range? \n"); 
    int rng2 = scanf("%d", &rng2); 
    printf("What would increment would you like to go up in? \n"); 
    int inc = scanf("%d", &inc); 

    do 
     { 
     printf("%d\n"rng1); 
     rng1 += inc; 
     } 
    while(rng1 != rng2); 
    } 
    return 0; 
    } 

从这个代码，我希望第一范围和第二范围在一定数量的上升之间的号码清单，而是我得到1.我做错了的值？ P.S.我试图'调试'它，发现当我用：

if(isalpha(rng1)); 
    printf("I am a String...") 
if(isdigit(rng1)) 
    printf("I am a Digit") 

它返回，“我是一个字符串...”。

谢谢！

Answer 1

你是通过调用scanf并且scanf已经将输入值分配给变量rng1,rng2和inc，并且scanf返回成功填充的参数列表的项目数量。因此，将scanf的返回值分配给这些变量是不正确的。只需使用输入数量的返回值即可。您应该读取值1，因为您只想读取每个scanf的一个值。此外，您还可以检查输入值以检测输入值是否有效。

除此之外，我想对您的代码进行一些修改。尤其是您的do{...}while();循环可能由于运算符!=而无限期地运行。请参阅下面的代码中的评论。

int main() 
{ 
    /* Declare the variables and do not assign the return value of scanf */ 
    int rng1, rng2, inc; 
    printf("Welcome to my number generator! \n"); 
    printf("What is the first number in the range? \n"); 
    /* repeat this check condition for each scanf, exit(EXIT_FAILURE) requires #include <stdlib.h>*/ 

    if (1 != scanf("%d", &rng1)) { 
     exit(EXIT_FAILURE); 
    } 
    printf("What is the second number in the range? \n"); 
    scanf("%d", &rng2); 
    printf("What would increment would you like to go up in? \n"); 
    scanf("%d", &inc); 

    do 
    { 
     printf("%d\n",rng1); 
     rng1 += inc; 
    } 
    /* Use <= instead of != and think about the case for rng1 is 1 rng2 is 5 and inc is 3, can you detect the end of the loop by adding 3 to the starting point 1 ? */ 
    while(rng1 <= rng2); 

    /* Remove the `}` here */ 
    return 0; 
} 

Answer 2

当您传递变量的地址（使用&运算符）时，允许scanf将扫描值写入变量。 scanf的返回值是而不是您正在查找的值（这是别的）。

因此，当您将scanf的返回值分配给您的变量时，您将覆盖它已有的正确值。这应该工作：

int rng1; 
scanf("%d", &rng1); 

Answer 3

的scanf()返回的值应始终进行检查，以确保手术成功。 格式字符串应该（几乎）总是包含一个前导空格“” 这样一个遗留的白色空间（如新行）被跳过/消耗 因此，代码应该更像：

int rng1; 
if(1 != scanf(" %d", &rng1)) 
{ // then, scanf failed 
    perror("scanf for rng1 failed"); 
    exit(EXIT_FAILURE); 
} 

// implied else, scanf for rng1 successful 

Answer 4

AS上面两个注释解释了scnaf（）有返回值只是为了检查成功的操作。然而，你需要阅读的价值是作为第二个参数。 我也可以看到代码中的括号不匹配。

int main(int argc, char **argv) 
{ 
    printf("Welcome to my number generator! \n"); 
    printf("What is the first number in the range? \n"); 
    int rng1; 
    int t = scanf("%d", &rng1);     // First scanned value stored in rng1 variable 
    printf("What is the second number in the range? \n"); 
    int rng2; 
    int t1 = scanf("%d", &rng2);     // Second scanned value stored in rng2 variable 
    printf("What would increment would you like to go up in? \n"); 
    int inc; 
    int t2 = scanf("%d", &inc);     // Third scanned value stored in inc variable 

    do 
    { 
     printf("%d\n",rng1); 
     rng1 += inc; 
    } 
    while(rng1 <= rng2);    // repeat until rng1 reaches value of rng2 

    return 0; 
} 

这里使用临时变量t，t1和t2，您可以验证值的扫描是否成功完成。