- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
void fillQueueWithStudents(Queue *q)
{
int counter = 1;
char answer = 'Y';
//this signifies the current variables that
//are being dealt with
char *currName;
double GPA;
student stud;
while((answer != 'N') || (answer != 'n'))
{
printf("Please enter the info for student %i \n",counter);
//Gets the name of the current student being entered
printf("Name: ");
scanf("%s", currName);
printf("\n");
//Gets the gpa of the current student being entered
printf("GPA: ");
scanf("%lf", &GPA);
//somewhere here is the problem
printf("\n");
//makes our student have the current variables
stud.name = currName;
stud.gpa = GPA;
//enqueues this student
enQueue(q,stud);
//increments to the next student
counter++;
//This sees if the user wants to enter another student
printf("Do you want to enter another student. Enter 'N' or 'n' for
no. \n");
scanf("%c", &answer);
printf("\n");
}
}
我在 scanf 之前和之后打印了 hello,它们都打印了。它不允许我输入 GPA 值。我想知道它是否正在扫描整个换行符或类似的东西。我不明白为什么它不让我输入一个值。我猜它试图在 GPA 变量中投入太多。
下面是完整的程序。
#include <stdlib.h>
#include <stdio.h>
typedef struct
{
char *name;
float gpa;
}student;
//This struct will represent a node in the queue
typedef struct Node
{
student *stud;
struct Node *next;
}QNode;
//This struct will identify the queue
//It will store the front and rear values to make
//it easier to enqueue and dequeue
typedef struct
{
QNode *front, *rear;
}Queue;
/*
This creates a new node with some data k.
It then sets its next value to NULL.
Then it returns the new node
Note: This node is not yet known to the queue. I will
use a separate function to do that
*/
QNode* newNode (student *stud)
{
QNode *temp = (QNode*)malloc(sizeof(QNode*));
temp->stud->gpa = stud->gpa;
temp->stud->name = stud->name;
temp->next = NULL;
return temp;
}
//Creats a empty queue
Queue* createQueue()
{
Queue *q = (Queue*)malloc(sizeof(Queue*));
q->front = q->rear = NULL;
return q;
}
/*
Uses the newNode function to make a new node in our linked list.
Then it puts that node in the correct place in the queue. This
is at the back
*/
void enQueue(Queue *q, student stud)
{
QNode *temp = newNode(&stud);//this creats a new node
/*
Checks if queue is empty, and if it is sets
our new node temp to rear and front. Then it
ends the function
*/
if(q->rear == NULL)
{
q->front = q->rear = temp;
return;
}
//sets the current rears next to our new node temp
q->rear->next = temp;
//this makes temp the new rear
q->rear = temp;
}
/*
This function serves/dequeues a node.
As always nodes are dequeued from the front of the
queue
*/
QNode *deQueue(Queue *q)
{
//if the queue is empty tell the user by
//returning NULL
if(q->rear == NULL)
{
return NULL;
}
/*
I will return front to the user in case they want
to use the data included in it. I have to save
front first to do that. So I save it hear
*/
QNode *temp = q->front;
//Now I make front equal to its next
q->front = q->front->next;
/*
This makes rear NULL if front is NULL
This means the list is empty and we have dequeued the last
node.
*/
if(q->front == NULL)
{
q->rear == NULL;
}
printf("The element dequeued was %s \n",temp->stud->name);
//This returns the node we have dequeued
return temp;
}
void printQueue(Queue *q)
{
//Check is the queue is empty and if it is stop the
//function
if(q->rear == NULL)
{
return;
}
/*
I will now print out each node out one by one.
To do this I create a QNode pointer and make it equal to the
front. I will then print all contents of front
and make temp equal to the next print and repeat this
till the end.
*/
QNode *temp = q->front;
while(temp != NULL)
{
int counter = 1;
printf("Student %i: Name: %s GPA: %f \n",
counter, temp->stud->name, temp->stud->gpa);
//increment the counter to display next student
counter++;
//Get the next node
temp = temp->next;
}
}
/*
This function keeps adding students to the queue until the
user enters 'N' that signifies they are done
*/
void fillQueueWithStudents(Queue *q)
{
int counter = 1;
char answer = 'Y';
//this signifies the current variables that
//are being dealt with
char *currName;
float GPA;
student stud;
while((answer != 'N') || (answer != 'n'))
{
printf("Please enter the info for student %i \n",counter);
//Gets the name of the current student being entered
printf("Name: ");
scanf("%s", currName);
printf("\n");
//Gets the gpa of the current student being entered
printf("GPA: ");
scanf("%f", &GPA);
printf("\n");
//makes our student have the current variables
stud.name = currName;
stud.gpa = GPA;
//enqueues this student
enQueue(q,stud);
//increments to the next student
counter++;
//This sees if the user wants to enter another student
printf("Do you want to enter another student. Enter 'N' or 'n' for no. \n");
scanf("%c", &answer);
printf("\n");
}
}
int main()
{
Queue *q = createQueue();
fillQueueWithStudents(q);
QNode *temp;
do {
temp = deQueue(q);
free(temp);
printQueue(q);
} while(temp != NULL);
return 0;
}
最佳答案
这个:
scanf("%s", currName);
是未定义的行为,因为 currName
未初始化。使用 scanf()
和普通 %s
来读取名称并不是一个好主意。使用缓冲区 (!),然后告诉 scanf()
缓冲区大小:
char currName[128];
if(scanf("%127s", currName) == 1)
{
}
...并检查返回值,如图所示。
请注意,%s
在第一个空白字符处停止,因此您将无法读取包含空格的名称。
关于c - 如果我使用 GPA 指针,为什么在使用 scanf ("%lf", GPA) 时会出现段错误,如果不是指针,则在使用 scanf ("%lf", &GPA) 时会出现段错误?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42955856/
我刚接触 C 语言几周,所以对它还很陌生。 我见过这样的事情 * (variable-name) = -* (variable-name) 在讲义中,但它到底会做什么?它会否定所指向的值吗? 最佳答案
我有一个指向内存地址的void 指针。然后,我做 int 指针 = void 指针 float 指针 = void 指针 然后,取消引用它们以获取值。 { int x = 25; vo
我正在与计算机控制的泵进行一些串行端口通信,我用来通信的 createfile 函数需要将 com 端口名称解析为 wchar_t 指针。 我也在使用 QT 创建一个表单并获取 com 端口名称作为
#include "stdio.h" #include "malloc.h" int main() { char*x=(char*)malloc(1024); *(x+2)=3; --
#include #include main() { int an_int; void *void_pointer = &an_int; double *double_ptr = void
对于每个时间步长,我都有一个二维矩阵 a[ix][iz],ix 从 0 到 nx-1 和 iz 从 0 到 nz-1。 为了组装所有时间步长的矩阵,我定义了一个长度为 nx*nz*nt 的 3D 指针
我有一个函数,它接受一个指向 char ** 的指针并用字符串填充它(我猜是一个字符串数组)。 *list_of_strings* 在函数内部分配内存。 char * *list_of_strings
我试图了解当涉及到字符和字符串时,内存分配是如何工作的。 我知道声明的数组的名称就像指向数组第一个元素的指针,但该数组将驻留在内存的堆栈中。 另一方面,当我们想要使用内存堆时,我们使用 malloc,
我有一个 C 语言的 .DLL 文件。该 DLL 中所有函数所需的主要结构具有以下形式。 typedef struct { char *snsAccessID; char *
我得到了以下数组: let arr = [ { children: [ { children: [], current: tru
#include int main(void) { int i; int *ptr = (int *) malloc(5 * sizeof(int)); for (i=0;
我正在编写一个程序,它接受一个三位数整数并将其分成两个整数。 224 将变为 220 和 4。 114 将变为 110 和 4。 基本上,您可以使用模数来完成。我写了我认为应该工作的东西,编译器一直说
好吧,我对 C++ 很陌生,我确定这个问题已经在某个地方得到了回答,而且也很简单,但我似乎找不到答案.... 我有一个自定义数组类,我将其用作练习来尝试了解其工作原理,其定义如下: 标题: class
1) this 指针与其他指针有何不同?据我了解,指针指向堆中的内存。如果有指向它们的指针,这是否意味着对象总是在堆中构造? 2)我们可以在 move 构造函数或 move 赋值中窃取this指针吗?
这个问题在这里已经有了答案: 关闭 11 年前。 Possible Duplicate: C : pointer to struct in the struct definition 在我的初学者类
我有两个指向指针的结构指针 typedef struct Square { ... ... }Square; Square **s1; //Representing 2D array of say,
变量在内存中是如何定位的?我有这个代码 int w=1; int x=1; int y=1; int z=1; int main(int argc, char** argv) { printf
#include #include main() { char *q[]={"black","white","red"}; printf("%s",*q+3); getch()
我在“C”类中有以下函数 class C { template void Func1(int x); template void Func2(int x); }; template void
我在64位linux下使用c++,编译器(g++)也是64位的。当我打印某个变量的地址时,例如一个整数,它应该打印一个 64 位整数,但实际上它打印了一个 48 位整数。 int i; cout <<
我是一名优秀的程序员,十分优秀!