将结构指针传递给C中的函数

Question

我无法将结构传递给一个以结构指针作为参数的函数,并且不断收到错误"错误:一元的无效类型参数'*'(有'StackNode')"

这是我的代码的必要部分(不是全部):

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

struct stackNode{
    char data;
    struct stackNode *nextPtr;
};
typedef struct stackNode StackNode;
typedef StackNode *StackNodePtr;

void convertToPostfix(char infix[], char postfix[]);
int isOperator(char c);
int precedence(char operator1, char operator2);
void push(StackNodePtr *topPtr, char value);
char pop(StackNodePtr *topPtr);
char stackTop(StackNodePtr topPtr);
int isEmpty(StackNodePtr topPtr);
void printStack(StackNodePtr topPtr);

int main(){
    convertToPostfix(NULL, NULL);
    return 0;
}

void convertToPostfix(char infix[], char postfix[]){
    StackNode stack = {'(', NULL};
    push(*stack, 'a');
    printStack(&stack);
}

void push(StackNodePtr* topPtr, char value){
    topPtr->nextPtr = NULL; //just temporary, not needed
    topPtr->data = value; //just temporary, not needed
}

任何帮助将不胜感激,谢谢

Answer 1

更改push通话,push(*stack, 'a'):

push(&stack, 'a');
     ^

间接on stack(*)没有意义,因为它不是指针.取其地址(&)的确如此.