一招搞定“C语言声明式”类型的面试题-白红宇

一招搞定“C语言声明式”类型的面试题

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发布时间：2019-02-26

本文共 3104 字，大约阅读时间需要 10 分钟。

C????????????????????????????????????????????????C??????????????????

C?????????

C?????????????????????????????????????????????????????????????????????????????????????????

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??*?????

const?volatile???????????int?long????????????????????

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????const?volatile???????????

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??1?char * const * p;

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p???????????

???????????char??????

p??????????????????

??2?char (* c[10])(int **p);

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c?????10???????

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???????int????????char???

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????????????????????????????cdecl.c????C????????????????????????????????????

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#include 
   
    #include 
    
     #include 
     
      #include 
      
       #define MAXTOKENS 100#define MAXTOKENLEN 64enum type_tag { IDENTIFIER, QUALIFIER, TYPE };struct token { char type; char string[MAXTOKENLEN]; };int top = -1;struct token stack[MAXTOKENS];struct token this;#define pop stack[--top]#define push(s) stack[++top] = svoid gettoken() {    char *s = this.string;    while ((*s = getchar()) == ' ') {        if (feof(stdin)) { *s = '\0'; break; }    }    if (isalnum(*s)) {        push(this);        while (isalnum(*s = getchar())) {            *s = '\0';        }        ungetc(*s, stdin);        this.type = classify_string();        return;    }    if (*s == '*') {        strcpy(this.string, "pointer to");        this.type = '*';        return;    }    this.string[1] = '\0';    this.type = *s;    return;}void read_to_first_identifier() {    gettoken();    while (this.type != IDENTIFIER) {        push(this);        gettoken();    }    printf("%s is ", this.string);    gettoken();}void deal_with_arrays() {    while (this.type == '[') {        printf("array ");        gettoken();        if (isdigit(this.string[0])) {            printf("0..%d ", atoi(this.string) - 1);            gettoken();        }        gettoken();        printf("of ");    }}void deal_with_function_args() {    while (this.type != ')') {        gettoken();    }    gettoken();    printf("function returning ");}void deal_with_pointers() {    while (stack[top].type == '*') {        printf("%s ", pop.string);    }}void deal_with_declarator() {    switch (this.type) {        case '[': deal_with_arrays(); break;        case '(': deal_with_function_args(); break;    }    deal_with_pointers();    while (top > 0) {        if (stack[top].type == '(') {            pop;            gettoken();            deal_with_declarator();        } else {            printf("%s ", pop.string);        }    }}int main() {    read_to_first_identifier();    deal_with_declarator();    printf("\n");    return 0;}

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char * const * p;char (* c[10])(int **p);

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p is pointer to function returning pointer to charc is array of 10 pointers to function returning pointer to char, function takes pointer to pointer to int and returns pointer to char