c - 从结构指针打印

Question

我已经使用 typedef struct 定义了一个结构然后在函数中创建一个指针。我将此指针传递给第二个函数以将值分配给结构元素。

typedef struct 
{
    double velocity; // there are more variables than this but its the 
                     // same for each variable so only showing it for one.
}Variables;

void information(Variables **constants);


int main()
{
double k;


Variables *Constants=NULL; // the structure variable

information(&Constants); // passed into the 'filling' function


k=Constants->velocity;
printf("Velocity %lf\n",k); //this displays correctly

printf("Velocity %lf\n",Constants->velocity); // this does not display correctly

return;
}

 void information(Variables **Constants)
{
 Variables *consts,constants; //creates the struct to be filled
consts=&constants; 

 constants.velocity=30.2;

*Constants=consts;  //assigns the pointer to the pointer passed into the function
return;

}

在这里你可以看到我显示了两次速度。第一次我将指针中的值分配给变量，一切都运行良好。如果我尝试使用 printf("Velocity %lf\n",Constants->velocity); 行直接显示该代码给出一个随机数。

我已使用 .dot 显示数字之前的结构格式，但从未通过指针，所以我不明白可能出了什么问题。

Answer 1

因为促销。它与 struct 或指针无关。

插图:

double k;
int i;

i = 123456;
k = i;       // here i is promoted to double before the assignment
             // takes place, now k contains 123456.0000

printf("Velocity %lf\n", k);  // this displays correctly
                              // because k is a double and %lf is the format
                              // specifier for double

printf("Velocity %lf\n", i);  // this does not display correctly
                              // because i is an int and if you use the %lf
                              // format specifier with an int, then you will
                              // get undefined behaviour hence the garbage
                              // that is printed

printf("Velocity %lf\n", (double)i);
                              // this does not display correctly because the cast
                              // converts the int to a double