把用空格分隔的数字到一个数组 [英] Putting numbers separated by a space into an array
问题描述
我想有一个用户输入一个空格分隔的数字，然后每个值存储为一个数组的元素。目前，我有：
I want to have a user enter numbers separated by a space and then store each value as an element of an array. Currently I have:
while ((c = getchar()) != '\n')
{
if (c != ' ')
arr[i++] = c - '0';
}
不过，当然，这个存储一个每个元素的数字。
but, of course, this stores one digit per element.
如果用户是键入：
10 567 92 3
我想10存储在改编[0] ，然后567 改编[1] 等。
我应该使用 scanf函数而不是以某种方式？
Should I be using scanf instead somehow?
推荐答案
有几种方法，这取决于您希望如何健壮的code是。
There are several approaches, depending on how robust you want the code to be.
最简单的就是使用 scanf函数与％d个转换说明：
The most straightforward is to use scanf with the %d conversion specifier:
while (scanf("%d", &a[i++]) == 1)
/* empty loop */ ;
的％d个转换符告诉 scanf函数来跳过任何前导空格和读取到下一个非数字字符。返回值是成功转换和分配的数量。由于我们正在阅读一个整数值，则返回值应该是1成功。
The %d conversion specifier tells scanf to skip over any leading whitespace and read up to the next non-digit character. The return value is the number of successful conversions and assignments. Since we're reading a single integer value, the return value should be 1 on success.
由于写的，这有一些陷阱。首先，假设你的用户输入更多的数字比你的数组的大小持有;如果你幸运的话，你会立即收到一条访问冲突。如果你没有，你会风重挫重要的事以后会引起问题（缓冲区溢出是一种常见的恶意软件攻击）。
As written, this has a number of pitfalls. First, suppose your user enters more numbers than your array is sized to hold; if you're lucky you'll get an access violation immediately. If you're not, you'll wind up clobbering something important that will cause problems later (buffer overflows are a common malware exploit).
所以，你至少要添加code，以确保你不走过去的数组的末尾：
So you at least want to add code to make sure you don't go past the end of your array:
while (i < ARRAY_SIZE && scanf("%d", &a[i++]) == 1)
/* empty loop */;
好为止。但现在假设您的用户fatfingers一个非数字字符在他们的输入，如 12 3r5 67 。按照规定，循环将分配 12 到 A [0] ， 3 为 A [1] ，那么将看到研究输入流中，返回0，不保存任何退出A [2] 。这里就是一个微妙的错误毛骨悚然的 - 即使没有被分配到 A [2] ，前pression 我++ 仍然得到评估，因此，您的认为的您分配东西<code> A [2] 即使它包含一个垃圾值。所以，你可能要推迟递增 I ，直到你知道你有一个成功读取：
Good so far. But now suppose your user fatfingers a non-numeric character in their input, like 12 3r5 67. As written, the loop will assign 12 to a[0], 3 to a[1], then it will see the r in the input stream, return 0 and exit without saving anything to a[2]. Here's where a subtle bug creeps in -- even though nothing gets assigned to a[2], the expression i++ still gets evaluated, so you'll think you assigned something to a[2] even though it contains a garbage value. So you might want to hold off on incrementing i until you know you had a successful read:
while (i < ARRAY_SIZE && scanf("%d", &a[i]) == 1)
i++;
在理想情况下，你想拒绝 3r5 干脆。我们可以读出的字符立即数以下，并确保它是空白;如果它不是，我们拒绝输入：
Ideally, you'd like to reject 3r5 altogether. We can read the character immediately following the number and make sure it's whitespace; if it's not, we reject the input:
#include <ctype.h>
...
int tmp;
char follow;
int count;
...
while (i < ARRAY_SIZE && (count = scanf("%d%c", &tmp, &follow)) > 0)
{
if (count == 2 && isspace(follow) || count == 1)
{
a[i++] = tmp;
}
else
{
printf ("Bad character detected: %c\n", follow);
break;
}
}
如果我们得到两个成功的转换，我们确保遵循是一个空白字符 - 如果不是这样，我们打印错误，并退出循环。如果我们得到1成功转换，这意味着有下面输入的号码没有字符（这意味着我们的数字输入后打EOF）。
If we get two successful conversions, we make sure follow is a whitespace character - if it isn't, we print an error and exit the loop. If we get 1 successful conversion, that means there were no characters following the input number (meaning we hit EOF after the numeric input).
另外，我们可以看到每个输入值的文本，并使用与strtol 来进行转换，这也可以让你赶上同一类问题（我的preferred法）：
Alternately, we can read each input value as text and use strtol to do the conversion, which also allows you to catch the same kind of problem (my preferred method):
#include <ctype.h>
#include <stdlib.h>
...
char buf[INT_DIGITS + 3]; // account for sign character, newline, and 0 terminator
...
while(i < ARRAY_SIZE && fgets(buf, sizeof buf, stdin) != NULL)
{
char *follow; // note that follow is a pointer to char in this case
int val = (int) strtol(buf, &follow, 10);
if (isspace(*follow) || *follow == 0)
{
a[i++] = val;
}
else
{
printf("%s is not a valid integer string; exiting...\n", buf);
break;
}
}
的别急，还有更多！的
假设你的用户是谁喜欢在你的code只是为了看看会发生什么，进入一个像扔厌恶那些输入扭曲QA类型之一<$c$c>123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890这是的显然的过大，以适应任何标准的整数类型。信不信由你， scanf函数（％d个，＆放大器; VAL）不会牦牛这一点，并且将结束存储的的东西到 VAL ，但同样是你可能想直接拒绝输入。
Suppose your user is one of those twisted QA types who likes to throw obnoxious input at your code "just to see what happens" and enters a number like 123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890 which is obviously too large to fit into any of the standard integer types. Believe it or not, scanf("%d", &val) will not yak on this, and will wind up storing something to val, but again it's an input you'd probably like to reject outright.
如果你只允许每行一个值，这成为比较容易防范; 与fgets 将换行符存储到目标缓冲区，如果有足够的空间，因此，如果我们看不到输入缓冲区换行符，那么用户输入的东西是超过我们再prepared处理：
If you only allow one value per line, this becomes relatively easy to guard against; fgets will store a newline character in the target buffer if there's room, so if we don't see a newline character in the input buffer then the user typed something that's longer than we're prepared to handle:
#include <string.h>
...
while (i < ARRAY_SIZE && fgets(buf, sizeof buf, stdin) != NULL)
{
char *newline = strchr(buf, '\n');
if (!newline)
{
printf("Input value too long\n");
/**
* Read until we see a newline or EOF to clear out the input stream
*/
while (!newline && fgets(buf, sizeof buf, stdin) != NULL)
newline = strchr(buf, '\n');
break;
}
...
}
如果你想允许每行多个值，如'10 20 30'，那么这个有点难度。我们可以回到从输入读取单个字符，并在每一个（警告，未经测试）做一个全面的检查：
If you want to allow multiple values per line such as '10 20 30', then this gets a bit harder. We could go back to reading individual characters from the input, and doing a sanity check on each (warning, untested):
...
while (i < ARRAY_SIZE)
{
size_t j = 0;
int c;
while (j < sizeof buf - 1 && (c = getchar()) != EOF) && isdigit(c))
buf[j++] = c;
buf[j] = 0;
if (isdigit(c))
{
printf("Input too long to handle\n");
while ((c = getchar()) != EOF && c != '\n') // clear out input stream
/* empty loop */ ;
break;
}
else if (!isspace(c))
{
if (isgraph(c)
printf("Non-digit character %c seen in numeric input\n", c);
else
printf("Non-digit character %o seen in numeric input\n", c);
while ((c = getchar()) != EOF && c != '\n') // clear out input stream
/* empty loop */
break;
}
else
a[i++] = (int) strtol(buffer, NULL, 10); // no need for follow pointer,
// since we've already checked
// for non-digit characters.
}
欢迎到交互式输入的奇妙疲惫不堪向上世界C.
Welcome to the wonderfully whacked-up world of interactive input in C.
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