- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
代码来自 Absolute beginner's guide to C,我得到这个错误 ./BlackJack.c<41> : warning C4047: '==' : 'int' differes in levels of indirection from 'char [2]'
当我尝试在 Visual Studio 中编译它时。这是二十一点游戏的主要功能。
第 <41> 行是 if (ans == "H") {
main()
{
int numCards; /* Equals 52 at the beginneing */
int cards[52], playerPoints[2], dealerPoints[2], total[2];
/* For user Hit/Stand or Yes/No response */
do { initCardsScreen(cards, playerPoints, dealerPoints, total, &numCards);
dealerGetsCard(&numCards, cards, dealerPoints);
printf("\n");
playerGetsCard(&numCards, cards, playerPoints);
playerGetsCard(&numCards, cards, playerPoints);
do {
ans = getAns("Hit or stand (H/S)? ");
if (ans == "H") {
platerGetsCard(&numCards, cards, playerPoints);
}
}
while ( ans != 'S');
totalIt(playerPoints, total, PLAYER);
/* Player's total */
do {
dealerGetsCard(&numCards, cards, dealerPoints);
} while (dealerPoints[ACEHIGH] < 17);
/* 17: Dealer stop */
totalIt(dealerPoints, total, DEALER);
/* Dealer's total */
findWinner(total);
ans = getAns("\nPlay again (Y/N)? ");
} while (ans == 'Y');
return;
}
(更新):这是完整的代码。
#include <stdio.h>
#include <time.h>
#include <ctype.h>
#include <stdlib.h>
#define BELL '\a'
#define DEALER 0
#define PLAYER 1
#define ACELOW 0
#define ACEHIGH 1
int askedForName = 0;
/****************************
This program specific prototype
****************************/
void dispTitle(void);
void initCardsScreen(int cards[52], int playerPoints[2], int dealerPoints[2], int total[2], int * numCards);
int dealCard(int * numCards, int cards[52]);
void dispCard(int cardDrawn, int points[2]);
void totalIt(int points[2], int total[2], int who);
void dealerGetsCard(int *numCards, int cards[52], int dealerPoints[2]);
void playerGetsCard(int *numCards, int cards[52], int playerPoints[2]);
char getAns(char mesg[]);
char ans;
void findWinner(int total[2]);
main()
{
int numCards; /* Equals 52 at the beginneing */
int cards[52], playerPoints[2], dealerPoints[2], total[2]; /* For user Hit/Stand or Yes/No response */
do { initCardsScreen(cards, playerPoints, dealerPoints, total, &numCards);
dealerGetsCard(&numCards, cards, dealerPoints);
printf("\n");
playerGetsCard(&numCards, cards, playerPoints);
playerGetsCard(&numCards, cards, playerPoints);
do {
char ans = getAns("Hit or stand (H/S)? ");
if (ans == "H") {
playerGetsCard(&numCards, cards, playerPoints);
}
}
while ( ans != 'S');
totalIt(playerPoints, total, PLAYER);
/* Player's total */
do {
dealerGetsCard(&numCards, cards, dealerPoints);
} while (dealerPoints[ACEHIGH] < 17);
/* 17: Dealer stop */
totalIt(dealerPoints, total, DEALER);
/* Dealer's total */
findWinner(total);
ans = getAns("\nPlay again (Y/N)? ");
} while (ans == 'Y');
return;
}
void initCardsScreen(int cards[52], int playerPoints[2], int dealerPoints[2], int total[2], int *numCards)
{
int sub, val=1; /* This function's Work variables */
char firstName[15]; /* Holds user's first name */
*numCards = 52; /* Holds running total of number of cards */
for (sub = 0; sub <= 51; sub++) { /* Counts from 0 to 51 */
val = (val == 14) ? 1 : val; /* If val is 14 reset to 1 */
cards[sub] = val;
val++; }
for (sub = 0; sub <= 1; sub++) { /* Counts from 0 to 1 */
playerPoints[sub] = dealerPoints[sub] = total[sub]=0; }
dispTitle();
if (askedForName ==0) { /* Name asked for nly once */
printf("\nWhat is your first name? ");
scanf_s(" %s", firstName);
askedForName = 1; /* Don't ask prompt again */
printf("Ok, %s, get ready for casino action!\n\n", firstName);
getchar(); /* Discards newline. You can safely */
} /* ignore compiler warning here. */
return;
}
/*** This function gets a card for the player and updates the player's points. ***/
void playerGetsCard(int *numCards, int cards[52], int playerPoints[2])
{
int newCard;
newCard = dealCard(numCards, cards);
printf("You draw: ");
dispCard(newCard, playerPoints);
}
/*** This function gets a card for the dealer and updates the dealer's poimts. ***/
void dealerGetsCard(int *numCards, int cards[52], int dealerPoints[2])
{
int newCard;
newCard = dealCard(numCards, cards);
printf("The dealer draws: ");
dispCard(newCard, dealerPoints);
}
/*** This function gets a card from the deck and stores it in either the dealer's or the player's hold of cards ***/
int dealCard(int * numCards, int cards[52])
{
int cardDrawn, subDraw;
time_t t; /* Gets time for a random value */
srand((unsigned int)(time(&t))); /* Seeds random-number generator */
subDraw = (rand() % (*numCards)); /* From 0 to numcards */
cardDrawn = cards[subDraw];
cards[subDraw] = cards[*numCards -1]; /* Puts top card */
(*numCards); /* in place of drawn one */
return cardDrawn;
}
/*** Displays the last drawn card and updates points with it. ***/
void dispCard(int cardDrawn, int points[2])
{
switch (cardDrawn) {
case(11) : printf("%s\n", "Jack");
points[ACELOW] += 10;
points[ACEHIGH] += 10;
break;
case(12) : printf("%s\n", "Queen");
points[ACELOW] += 10;
points[ACEHIGH] += 10;
break;
case(13) : printf("%s\n", "King");
points[ACELOW] += 10;
points[ACEHIGH] += 10;
break;
default : points[ACELOW] += cardDrawn;
if (cardDrawn == 1)
{
printf("%s\n", "Ace");
points[ACEHIGH] += 11;
}
else
{
points[ACEHIGH] += cardDrawn;
printf("%d\n", cardDrawn);
}
} return;
}
/*** Figure the total for player or dealer to see who won. This function takes into account the fact that Ace is either 1 or 11. ***/
void totalIt(int points[2], int total[2], int who)
{
/* The following routine first looks to see if the total points counting Aces as 1 is equal to the total points couning Aces as 11. If so,
or if the total points counting Aces as 11 is more than 21, the program uses the total with Aces as 1 only */
if ((points[ACELOW] == points[ACEHIGH]) || (points[ACEHIGH] > 21)) {
total[who] = points[ACELOW]; /* Keeps all Aces as 1 */
}
else {
total[who] = points[ACEHIGH]; /* Keeps all Aces as 11 */
}
if (who == PLAYER) /* Determines the message printed */ {
printf("You have a total of %d\n\n", total[PLAYER]);
}
else {
printf("The house stands with a total of %d\n\n", total[DEALER]);
}
return;
}
/*** Prints the winning player. ***/
void findWinner(int total[2])
{
if (total[DEALER] == 21) {
printf("The house wins.\n");
return;
}
if ((total[DEALER] > 21) && (total[PLAYER] > 21)) {
printf("%s", "Nobody wins.\n");
return;
}
if ((total[DEALER] >= total[PLAYER]) && (total[DEALER] < 21)) {
printf("The house wins.\n");
return;
}
if ((total[PLAYER] > 21) && (total[DEALER] < 21)) {
printf("The house wins.\n");
return;
}
printf("%s%c", "You win!\n", BELL);
return;
}
/*** Gets the user's uppercase, single-character response. ***/
char getAns(char mesg[])
{
char ans;
printf("%s", mesg); /* Prints the prompt message passed */
ans = getchar();
getchar(); /* Discards newline. You can safely ignore compiler warning here. */
return toupper(ans);
}
/*** Clears everything off the screen. ***/
void dispTitle(void)
{
int i=0;
while (i < 25) { /* Clears screen by printing 25 blank lines to 'push off' stuff that
might be left over on the screen before this program */
printf("\n");
i++;
}
printf("\n\n*Step right up to the Blackjack tables*\n\n");
return;
}
最佳答案
请注意所有其他比较都是如何用单引号引起来的?
while (ans == 'Y')
在那里,您正在比较单个字符(这是一个数字,因此可以与另一个数字进行比较)。
另一方面,在失败的代码中,您使用了双引号。
ans == "H"
因此,您正在与一个字符串进行比较,它是一个字符数组。该数组有两个字符长以容纳 null terminator .
关于c - int 与 char[2] 错误的间接级别不同,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29676872/
这个问题在这里已经有了答案: Why don't Java's +=, -=, *=, /= compound assignment operators require casting? (11 个
当我尝试运行以下代码时,List(.of) 无法编译并给出主题错误。 package collections; import java.util.LinkedHashSet; import java.
我正在尝试编译使用 ChatScript 库的程序。这是我在名为 main.cpp 的文件中的代码: #include #include "common.h" using namespace std
我想在我的程序中外部使用 ChatScript。在documents它说: Embedding Step #1 First, you will need to modify `common.h and
假设我有一个 char,我想用一行代码将其 strcat() 转换为 char 数组。对于 [一个非实用的] 示例: strcat("ljsdflusdfg",getchar()); 或者我想做相反的
我有以下类型签名: *Main Lib> let f :: a -> a -> a -> a; f = undefined *Main Lib> let x :: Char; x = undefin
我正在学习如何在 C 中使用指针(使用 malloc 和 free),但我在这个练习中遇到了一些麻烦。我只想制作一个指针数组,我想在其中保存每个单词的方向。然后我想为一个特定的词做一个 free(),
我有一个字符*: char* version = "10.5.108"; 我想通过字符分隔符获取两个新的 char*。 char delimiter = '.'; 执行以下代码后: printf("|
最近在学习Cpp,今天在学习使用Clion做测试的时候,发生了奇怪的事情。 这是我的代码 int main() { char c = 'b'; char carr[1]{'a'};
我对 c 很陌生,我正在审查一些代码。我遇到了这个: static char * fromDataType; static char * toDataType; static char * fromR
我有一个像这样的动态结构: struct network { int count; char** ips; } 如果我知道每个字符串数组都是 16 个字节(即 INET_ADDRSTR
我有一个旧程序,其中使用了一些库函数,但我没有那个库。 所以我正在使用 C++ 库编写该程序。在那个旧代码中有一些函数是这样调用的 *string = newstrdup("这里有一些字符串"); 字
我正在编写一个函数,该函数接受 ArrayList,然后将每个 char[] 复制到另一个增加长度的 char[] 中,然后将新的 char[] 添加到新的 ArrayList 中。当我尝试复制数组时
我正在寻找 map >并生成每个可能的 map从它。 我知道这可能会占用大量内存并需要一些时间。 每个map需要包含每个字母 a-z,并映射到唯一的 a-z 字符。 IE。啊bjcp迪EVfh嘎血红蛋
#define NAME_LEN 20 #include "stdio.h" #include "stdlib.h" #include "string.h" #pragma warning(disab
所以我必须创建一个函数来找到一对带有第一个字母并返回第二个字母的函数。 我实际上找到了一个答案,但是使用 map 功能却找不到。 lookUp :: Char -> [(Char, Cha
我最近接受采访并要求写mystrcat(*s1, *s2, *s3) 其中s1 和s2 是源字符串连接结果由 s3 给出。有人告诉我,不要担心 s3 的内存分配,并假设 s1 和 s2 不是空/无效字
今天我与一位同事讨论了他(对我来说)不寻常的“main”函数签名。他喜欢这样声明: int main(int argc, char* (*argv)[]) { printf("at index
这个问题在这里已经有了答案: 关闭 12 年前。 Possible Duplicate: What's the difference between new char[10] and new cha
通常字符串文字是 const char[] 类型。但是当我把它当作其他类型时,我得到了奇怪的结果。 unsigned char *a = "\355\1\23"; 使用此编译器会抛出警告“初始化中的指
我是一名优秀的程序员,十分优秀!