c - 在 c 中使用未声明的标识符

Question

我在行中收到未声明的标识符错误:“printf("%f", new_u[i]);” 这很奇怪，因为我可以在那个 for 循环中打印 i 它有值。为什么会出现该错误？

const int MAX = 101;

int main(void) {

    int t = 1; //time
    int m = 0; //number of segments of bar
    int n = 0; //number of time intervals

    double new_u[MAX]; //to store temps currently being converted (array of 101 doubles)
    double old_u[MAX]; //to store temps corresponding to prev time (array of 101 doubles)

    printf("Enter number of segments: ");
    scanf("%d", &m);
    printf("Enter number of time intervals: ");
    scanf("%d", &n);

    double h = (1.0/m); //length of bar segments
    double d = (1.0/n); //length of time interval

    for (int j = 1; j <= n; j++) { //j is which time interval the iteration is on
        int t_j = j * d; //t_j is the actual fraction of a second the iteration is on (i.e. 0.0, 0.2, 0.4...)
        new_u[0] = new_u[m] = 0.0;
        for (int i = 1; i < m; i++)
            new_u[i] = old_u[i] + d/(h*h)*(old_u[i-1] - 2*old_u[i] + old_u[i+1]);
        printf("%f", new_u[i]);
        //I need to finish code by printing new_u values
        //Then copy new_u into old_u for next pass;
    }

}

Answer 1

因为您没有在内部 for 循环中使用任何大括号，所以 i 的值对于 printf 语句 是未知的。在条件语句和循环语句中，如果没有大括号(或创建的 block )用于它们，那么它们只能操作并且范围仅限于声明之后的语句。

for (int i = 1; i < m; i++)
        new_u[i] = old_u[i] + d/(h*h)*(old_u[i-1] - 2*old_u[i] + old_u[i+1]); //i  is known here
        printf("%f", new_u[i]); //i is not available for this

像这样使用大括号

 for (int i = 1; i < m; i++)
 {
    new_u[i] = old_u[i] + d/(h*h)*(old_u[i-1] - 2*old_u[i] + old_u[i+1]);
    printf("%f", new_u[i]);
 }