如何修复我的代码以读取函数中的字符串，该函数返回指向结构的指针 [英] How to fix my code to read a string in a function that returns a pointer to a struct

问题描述

我正在编写一个代码，该代码使用一个函数来返回指向结构的指针，该结构是动态分配的.但是，我的代码没有读取字符串.当我运行它时，它只跳了类型名称"部分，键入了年龄，并且它显示了年龄，并且没有显示任何名称..............................................................................................奇怪的是，当我使用scanf读取字符串时，该代码有效，但gets或fgets却没有.谁能帮我吗?预先感谢.

I'm writing a code that uses a function to return a pointer to a struct, which is allocated dynamically. However, my code isn't reading strings. When I run it, it simply jumps the "Type name" part, I type the age, and it prints the age and nothing for the name. Strangely, the code works when I use scanf to read the string, but it didn't with gets or fgets. Could anyone please help me? Thanks in advance.

#include <stdio.h>
#include <stdlib.h>

struct details
{
    char name[100];
    int age;
};

struct details * details_pointer(int n)
{
    struct details *pointer = (struct details *) malloc (n*sizeof(struct details));
    for (int i=0; i<n; i++)
    {
        printf("Student %d:\n", i);
        printf("name:\n");
        scanf("%s", pointer[i].name);
        //gets(pointer[i].name); not working
        //fgets(pointer[i].name, 100, stdin); not working
        printf("age:\n");
        scanf("%d", &pointer[i]. age);
    }
    return pointer;
}

int main()
{
    int n;
    printf("Type the number of persons:\n");
    scanf("%d", &n);
    struct details *student = details_pointer(n);
    for (int i=0; i<n; i++)
    {
        printf("\nName: %s", (*(student+i)).name);
        printf("Age: %d\n", (*(student+i)).age);
    }
    free(student);
    system("pause");
    return 0;
}

推荐答案

这是因为scanf在输入流中留下了换行符. fgets在调用它时将其作为名称.为了证明这一点，请更改:

This is because scanf leaves a newline in the input stream. fgets gets it as the name when it is called. To prove this, change:

scanf("%d", &n);

类似于:

n = 1;

您将看不到任何问题.

如果您不想使用scanf，则可以先呼叫fgets，然后再呼叫atoi/strtol.

If you don't want to use scanf, you can call fgets then atoi/strtol.

char    *num;
fgets(num, 100, stdin);
n = atoi(num);

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