表达式语法错误 [英] error in expression syntax

问题描述

I was clear the first error "ne = (temp *) malloc(sizeof(temp)); //error in expression syntax". correct syntax is: ne = (struct node *) malloc(sizeof(struct node));

2) i am trying 2nd error "<b>printf("%d\n",front->data); // error in pointer structure required on left side of -> or->*</b>" any one help me how to clear this error

#include
#include

// creating structure
struct node
{
int data;
struct node *next;	// this variable is refering the struct node
}*front,*temp,*get,*ne;

int main()
{
front = NULL;
clrscr();
// insert the queue element
// insert function declaration
add(&front,56);
add(&front,88);
add(&front,44);
// display the queue element
display(front);

getch();
return 0;
}

//definition of the insert function
add(struct node *temp,int val)
{
ne = (temp *) malloc(sizeof(temp)); //error in expression syntax
//ne = (temp *) malloc(10*sizeof(temp)); //error in expression syntax
if(temp == NULL)
{
temp->data = val;
temp->next = NULL;

ne = temp;
}
else
//struct node *p;
{
get = front;
get->data = val;
get->next = ne;
}
}

display(front)
{
int count=0;
while(front != NULL)
{
printf("%d\n",front->data); // error in pointer structure required on left side of -> or->*
//printf("%d\n",(*front).data); // error in invalid inderiction
count++;
}
printf("No of Nodes: %d\n",count);
}

推荐答案

在

display(front)

中，您尚未声明，因此编译器假定它是int.然后，您不能期望它充当->左侧的指针.您需要用

you haven''t declared a type for front, so the compiler assumes it is an int. You can''t then expect it to behave as a pointer on the left side of ->. What you need is to replace that line with

display(struct node *front)

替换该行，然后您的printf()调用中的任何一个将编译.

彼得

Then either of your printf() calls will compile.

Peter

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