/* This is a code that changes infix notation to postfix notation. I used FILE/IO to get infix notations and the infix.txt file looks like
/ *这是将中缀表示法更改为后缀表示法的代码。我用FILE / IO来获取中缀符号，infix.txt文件看起来像
3
2+4*2-1;
9+3^2^(3-1)*2;
2*((7-2)/3+4)^2%3;
My question is I get error saying "control may reach end of non-void function" on the last two functions, icp and isp. How can I fix this? */
我的问题是我在最后两个函数icp和isp上得到错误说“控制可能达到非空函数的结束”。我怎样才能解决这个问题？ * /
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>
#define MAX_SIZE 100
#define ENTER 0x000d
void infixtopostfix(char expression[]);
char get_token(char expression[], int *index);
void push(int *top, char token);
int pop(int *top);
int icp(char op);
int isp(char op);
char stack[MAX_SIZE];
int main(void) {
int i, num;
FILE *file;
char expression[MAX_SIZE];
if((file=fopen("infix.txt","r")) == NULL) {
printf("No file.\n");
}
fgets(expression, MAX_SIZE, file);
num = atoi(expression);
for(i=0; i < num; i++) {
fgets(expression, MAX_SIZE, file);
printf("%s", expression);
infixtopostfix(expression);
}
fclose(file);
}
void infixtopostfix(char expression[]) {
char token;
char element;
int top=0;
int index=0;
for(token=get_token(expression, &index); token!=';'; token=get_token(expression, &index)) {
if(isdigit(token)) printf("%c", token);
else if(token == ')') {
while (stack[top] != '(') {
element = pop(&top);
printf("%c", element);
}
pop(&top);
}
else {
while (isp(stack[top])>=icp(token)) {
element = pop(&top);
printf("%c", element);
}
push(&top, token);
}
}
while((token=pop(&top))!=0) printf("%c", token);
printf("\n");
}
char get_token(char expression[], int *index) {
char token=expression[(*index)++];
return token;
}
void push(int *top, char data) {
if(*top < (MAX_SIZE-1)) stack[++(*top)] = data;
}
int pop(int *top) {
if(*top > -1) return stack[(*top)--];
else return 0;
}
int icp(char op) {
switch (op){
case '(' : return 20; break;
case '+' : return 12; break;
case '-' : return 12; break;
case '*' : return 13; break;
case '%' : return 13; break;
case ';' : return 0;
}
}
int isp(char op) {
switch (op){
case '(' : return 0; break;
case '+' : return 12; break;
case '-' : return 12; break;
case '*' : return 13; break;
case '%' : return 13; break;
case ';' : return 0;
}
}
2 个解决方案
#1
2
While you may believe that your methods will only ever get one of these 6 characters, the compiler has to assume that any character can be passed in. At the moment, these methods will not return anything if you pass in, for example, 'x'.
虽然您可能认为您的方法只能获得这6个字符中的一个，但编译器必须假定可以传入任何字符。目前，如果传入，这些方法将不会返回任何内容，例如，'x ”。
You need to specify a default action for all characters not otherwise covered by your switch. Whether this is returning a default value or throwing an exception, that is up to you. But it's nonetheless a possibility your code needs to handle.
您需要为交换机未覆盖的所有字符指定默认操作。无论是返回默认值还是抛出异常，都取决于您。但它仍然是您的代码需要处理的可能性。
#2
0
like this:
喜欢这个：
}//end switch
fprintf(stderr, "\nunrecognized OP(%c) is specified.\n", op);// '/', '^'...
return -1;//or exit(-1);
#1
2
While you may believe that your methods will only ever get one of these 6 characters, the compiler has to assume that any character can be passed in. At the moment, these methods will not return anything if you pass in, for example, 'x'.
虽然您可能认为您的方法只能获得这6个字符中的一个，但编译器必须假定可以传入任何字符。目前，如果传入，这些方法将不会返回任何内容，例如，'x ”。
You need to specify a default action for all characters not otherwise covered by your switch. Whether this is returning a default value or throwing an exception, that is up to you. But it's nonetheless a possibility your code needs to handle.
您需要为交换机未覆盖的所有字符指定默认操作。无论是返回默认值还是抛出异常，都取决于您。但它仍然是您的代码需要处理的可能性。
#2
0
like this:
喜欢这个：
}//end switch
fprintf(stderr, "\nunrecognized OP(%c) is specified.\n", op);// '/', '^'...
return -1;//or exit(-1);