如何从C中的结构打印指针变量？

Question

I need to print out a pointer variable within a structure. I assume i must de-reference but am not sure how without getting a segmentation fault.

struct HealthProfile{ //structure with pointers to all needed variables.
    char *name;
    char *last;
    char *gender;
    struct date *dob;
    float *height;
    float *weight;
};

void readData(){
    float height;
    printf("What is your name?\n");
    scanf("%s", &H.name); //scan
    //H.name = name;
    printf("What is your last name? \n");
    scanf("%s", &H.last);
    //H.last = last;
    printf("What is your Height name? \n");
    scanf("%f", &H.height);


    printf("Height: %f\n", *(H.height));
    //printf("First Name: %s\n", H->name);
    //printf("Last Name: %s\n", H->last);
}

I want it to print out the scanned in height which is a float.

Answer 1

First of all, instead of declaring float height; , you need to declare struct HealthProfile H; . Better yet, declare struct HealthProfile profile; and replace H everywhere with profile .

Next, fix your scanf() statements. For example

scanf("%s", &H.name);

should be

scanf("%s", profile.name);

Similarly change

scanf("%f", &H.height);

to

scanf("%f", profile.height);

Now your syntax for printf() will be correct.

However, you will still have problems because non of your pointers have been allocated memory. Declaring the name and last fields as pointers makes sense. However, I think you should declare float height; and float weight; instead of using pointers for these values. If you do so, then your original scanf() statements with the & operator will be correct.