#include <stdio.h>
int main(void) {
char *matchs[10] = { "2:3", "0:0", "15:12", "2:13", "1:5", "55:7", "7:2", "17:2",
"2:17", "17:18"};
char *(*pt) = matchs; // Pointing to array matchs
for(int i = 0; i < 10; i++){
printf("%s ", *++(pt));
}
return 0;
}
我们的足球队以十场比赛获得冠军。每个匹配的结果看起来像"；x： y"；。所有匹配的结果都记录在阵列中
例如：["3:1", "22:2", "0:19", ...]
写一个函数，用这样的列表来计算每支球队在锦标赛中的积分。每场比赛的点数规则：
if x>y - 3 points
if x<y - 0 point
if x = y - 1 points
我们不能使用stoi()或令牌函数，而只能使用指针，while循环和for循环提取数据。
这段代码是我尝试过的，但不起作用。
输出：0:0 15:12 2:13 1:5 55:7 7:2 17:2 2:17 17:18
以下是整个解决方案。你可以通过在每个结果中找到:字符来获得y队得分的指针。(请参阅如何获取str_y指针(。我认为最棘手的部分是在不使用stoi这样的函数的情况下获得每支球队的分数。我确实写了我的解决方案，但我建议你先自己写(尤其是如果这是家庭作业(。
#include <stdio.h>
int main(void)
{
int point_x = 0;
int point_y = 0;
const char *matches[10] = { "2:3", "0:0", "15:12", "2:13", "1:5", "55:7", "7:2", "17:2", "2:17", "17:18" };
for (int i = 0; i < 10; i++)
{
// This pointer points the first character of each result
const char* match = matches[i];
// Get the pointer of ":" of this result
const char* separator = match;
while( *separator != ':' && *separator != ' ' ) separator++;
// You would need to handle error when ':' is not found but here just assumes all the results contain ":".
// To make it easy to understand prepare two pointers to point each score in a result.
const char* str_x = match;
const char* str_y = separator+1; // next character of the separator
// Convert each string to number. Because we cannot use stoi, tokenize functions
// we have to do it by ourselves.
int score_x = 0;
while( *str_x != ':'){ // until separator
score_x = score_x*10 + (*str_x - '0');
str_x++;
}
int score_y = 0;
while( *str_y != ' '){ // until NULL
score_y = score_y*10 + (*str_y - '0');
str_y++;
}
// Now compare them and give points
if( score_x == score_y ){
point_x++;
point_y++;
}else if( score_x > score_y ){
point_x+=3;
}else{
point_y+=3;
}
}
printf("point_x=%d point_y=%dn", point_x, point_y);
return 0;
}
char*matches[10]是一个char指针数组或字符串数组。我们可以指向数组中具有matches[index]或*(matches+index(的任何一个字符串，这些字符串会衰减为char指针(字符串(。然后，我们可以通过[]或指针算术的解引用在其中找到一个特定的字符。
例如，
#include <stdio.h>
int main()
{
char *matches[10] = { "2:3", "0:0", "15:12", "2:13", "1:5", "55:7", "7:2", "17:2",
"2:17", "17:18"};
int i;
for(i = 0; i < 10; i++)
{
/*Dereference once will give a char pointer (string), dereference twice
will give us a single char*/
printf("n%s", *(matches + i)); //This will print the whole string, i.e. "2:3" or {2,:,3,0}
printf("nx: %c", *(*(matches + i) + 0)); //This will print '2'
printf("ny: %c", *(*(matches + i) + 2)); //This will print '3'
/*Note we can compare chars directly, as they are essentially 1 byte ints, so
('2' < '3') will evaluate to 1 (true)*/
/*Note we will also run into problems with scores that are multiple chars, such as 17:20
We can work around this with an if statement checking if a char is ':', which
divides the numbers. We can then print them out or write them to a file or
whatever. Note we can find a literal accurate integer representation of a char
without stoi or atoi by doing: int number = '9' - '0'; This will give the number 9.
*/
}
return 0;
}
编辑：为了详细说明关于计算多个字符的分数，
#include <stdio.h>
#include <math.h>
int str_to_int(const char * string, int length)
{
int i;
int num = 0;
for(i = 0; i < length; i++)
{
int place_value = pow(10, (length - i - 1));
num += place_value * (*(string + i) - '0');
}
return num;
}
int main()
{
char *matches[10] = { "2:3", "0:0", "15:12", "2:13", "1:5", "55:7", "7:2", "17:2",
"2:17", "17:18"};
int i;
for(i = 0; i < 10; i++)
{
/*Dereference once will give a char pointer (string), dereference twice
will give us a single char*/
printf("n%s", *(matches + i)); //This will print the whole string, i.e. "2:3" or {2,:,3,0}
char current; //To hold the current char we are reading.
int counter = 0;
char number[5]; //To hold a string representation of a number
do
{
current = *(*(matches + i) + counter);
if(current != ':')
*(number + counter) = current;
counter++;
}
while (current != ':');
*(number + counter - 1) = ' ';
//We now have the x value.
printf("nx: %d", str_to_int(number, counter - 1));
int counter2 = 0;
do
{
current = *(*(matches + i) + counter);
*(number + counter2) = current;
counter2++;
counter++;
}
while (current != ' ');
//Now we have y
printf("ny: %d", str_to_int(number, counter2 - 1));
}
return 0;
}
这将能够找到具有多个字符的分数。
编辑2：输出：
2:3
x: 2
y: 3
0:0
x: 0
y: 0
15:12
x: 15
y: 12
2:13
x: 2
y: 13
1:5
x: 1
y: 5
55:7
x: 55
y: 7
7:2
x: 7
y: 2
17:2
x: 17
y: 2
2:17
x: 2
y: 17
17:18
x: 17
y: 18
unsigned points = 0;
for(int i = 0; i < 10; i++){
int r;
char *itr;
for (itr = match[i]; itr != ':'; itr++);
r = strncmp(match[i], itr + 1, itr - match[i]);
if (!r)
points += 1;
else if (r > 0)
points += 3;
}
如果我们用指针对指针消除了所有不必要和奇怪的复杂性，我们就会得到这个快速可读的代码：
int main(void)
{
char* matchs[10] = { "2:3", "0:0", "15:12", "2:13", "1:5",
"55:7", "7:2", "17:2", "2:17", "17:18" };
for(int i = 0; i < 10; i++)
{
printf("%s ", matchs[i]);
}
return 0;
}
使用char**唯一有意义的情况是，字符串的数量未知，并且它们以sentinel值(如NULL(终止。这是一种比提前知道项目数量更不理想的情况，因此执行起来有些慢。示例：
#include <stdio.h>
int main(void)
{
char* matchs[10+1] = { "2:3", "0:0", "15:12", "2:13", "1:5",
"55:7", "7:2", "17:2", "2:17", "17:18", NULL };
for(char** ptr=matchs; *ptr != NULL; ptr++)
{
printf("%s ", *ptr);
}
return 0;
}
从那时起，如果您愿意，您可以将每个项目转换为整数，而不是printf调用。这是另一项任务，您可能需要使用strtoul。