如何在此堆栈中打印值？

Question

我找到了一些代码来实现堆栈的C实现,并决定使用它.但是,有几个typedef,我很难在stackT中打印值(实际上是一个char数组).下面是代码.我究竟做错了什么？

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

typedef char stackElementT;

typedef struct {
  stackElementT *contents;
  int maxSize;
  int top;
} stackT;

void StackInit(stackT *stackP, int maxSize) {
    stackElementT *newContents;
    newContents = (stackElementT *)malloc(sizeof(stackElementT)*maxSize);
    if (newContents == NULL) {
        fprintf(stderr, "Not enough memory.\n");
        exit(1);
    }

    stackP->contents = newContents;
    stackP->maxSize = maxSize;
    stackP->top = -1; //empty...
}

void StackDestroy(stackT *stackP) {
    free(stackP->contents);
    stackP->contents = NULL;
    stackP->maxSize = 0;
    stackP->top = -1; //empty
}

int StackIsEmpty(stackT *stackP) {
    return stackP->top < 0;
}

int StackIsFull(stackT *stackP) {
    return stackP->top >= stackP->maxSize-1;
}

void StackPush(stackT *stackP, stackElementT element) {
    if(StackIsFull(stackP)) {
        fprintf(stderr, "Can't push element: stack is full.\n");
        exit(1);
    }
    stackP->contents[++stackP->top] = element;
}

stackElementT StackPop(stackT *stackP) {
    if(StackIsEmpty(stackP)) {
        fprintf(stderr, "Can't pop element: stack is empty.\n");
        exit(1);
    }
    return stackP->contents[stackP->top--];
}

void StackDisplay(stackT *stackP) {
    if(StackIsEmpty(stackP)) {
        fprintf(stderr, "Can't display: stack is empty.\n");
        exit(1);
    }
    int i;
    printf("[ ");
    for (i = 0; i < stackP->top; i++) {
        printf("%c, ", stackP[i]); //the problem occurs HERE
    }
    printf("%c ]", stackP[stackP->top]);
}

int postfix(char* expr, int length) {
    int i;
    stackT stack;
    StackInit(&stack, 1000);
    int temp;
    for (i = 0; i < length; i++) {
        if ((expr[i] >= 48) && (expr[i] <= 57)) {
            printf("Is a number! Pushed %d\n", expr[i]);
            StackPush(&stack, expr[i]);
        }
        else {
            switch (expr[i]) {
                case 43: {
                    temp = StackPop(&stack);
                    StackPush(&stack, StackPop(&stack)+temp);
                }
                    break;
                case 45: {
                    temp = StackPop(&stack);
                    StackPush(&stack, StackPop(&stack)-temp);
                }
                    break;
                case 47: {
                    temp = StackPop(&stack);
                    StackPush(&stack, StackPop(&stack)/temp);
                }
                    break;
                case 42: {
                    temp = StackPop(&stack);
                    StackPush(&stack, StackPop(&stack)*temp);
                }
                    break;
                default:
                    break;
            }
        }
    }
    return StackPop(&stack);
}

int main() {
    int i;
    char* expr = "1 2 3 + * 3 2 1 - + *";
    for(i = 0; expr[i] != '\0'; i++) ;
    printf("%d\n", postfix(expr, i));
}

Answer 1

编译器(Mac OS X 10.6.7上的GCC 4.2.1)告诉我:

$ cc -O -std=c99 -Wall -Wextra     st.c   -o st
st.c: In function ‘StackDisplay’:
st.c:72: warning: format ‘%c’ expects type ‘int’, but argument 2 has type ‘stackT’
st.c:74: warning: format ‘%c’ expects type ‘int’, but argument 2 has type ‘stackT’
$

在我的代码版本中,这两行是printf()语句StackDisplay(),在您声明遇到问题的地方.

void StackDisplay(stackT *stackP)
{
    if(StackIsEmpty(stackP)) {
        fprintf(stderr, "Can't display: stack is empty.\n");
        exit(1);
    }
    int i;
    printf("[ ");
    for (i = 0; i < stackP->top; i++) {
        printf("%c, ", stackP[i]); //the problem occurs HERE
    }
    printf("%c ]", stackP[stackP->top]);
}

你可能想要stackP->contents[i].通过该修复,程序"运行"但产生:

Can't pop element: stack is empty.

现在,这是你要解决的问题.

(哦,我还在评论中诊断出的for循环后修复了杂散分号main().)

循环应该写成strlen(expr)(然后你需要#include <string.h>).实际上,主程序的主体简化为:

char* expr = "1 2 3 + * 3 2 1 - + *";
printf("%d\n", postfix(expr, strlen(expr)));

您通常应该将top索引保存到下一个要使用的位置,因此初始值通常0不是-1.

不要学习数字的ASCII代码 - 忘了你曾经做过.

    if ((expr[i] >= 48) && (expr[i] <= 57)) {

你应该写:

    if ((expr[i] >= '0') && (expr[i] <= '9')) {

或者,更好(但你必须#include <ctype.h>):

    if (isdigit(expr[i])) {

类似的评论适用于交换机:

        switch (expr[i]) {
            case 43: {
                temp = StackPop(&stack);
                StackPush(&stack, StackPop(&stack)+temp);
            }
                break;

我不知道缩进背后的逻辑,但43应写为'+',45为'-',47 '/'和42 '*'.

这会产生:

Is a number! Pushed 49
Is a number! Pushed 50
Is a number! Pushed 51
Is a number! Pushed 51
Is a number! Pushed 50
Is a number! Pushed 49
68

如果您修复了数字推送代码,如下所示:

printf("Is a number! Pushed %d\n", expr[i] - '0');
StackPush(&stack, expr[i] - '0');

然后你得到:

Is a number! Pushed 1
Is a number! Pushed 2
Is a number! Pushed 3
Is a number! Pushed 3
Is a number! Pushed 2
Is a number! Pushed 1
20

并使用更多的仪器,按照以下方式:

temp = StackPop(&stack);
printf("Sub: result %d\n", temp);
StackPush(&stack, temp);

每次操作后,结果是:

Is a number! Pushed 1
Is a number! Pushed 2
Is a number! Pushed 3
Add: result 5
Mul: result 5
Is a number! Pushed 3
Is a number! Pushed 2
Is a number! Pushed 1
Sub: result 1
Add: result 4
Mul: result 20
20

你很亲密