有帮助吗？ [英] Any help?
问题描述
有人可以帮我解决这个问题吗？我想我已经把它缩小了
到4中有2个但不确定。
char array [80] ="我很简短;
这是真的吗？数组的大小将减少到10或
数组大小保持在80并且字符串填充数组位置，从数组[0]开始是
。
cj写道：有人可以帮我解决这个问题吗？我想我已经把它缩小到4个中的2个但不确定。
char数组[80] ="我很矮;
这是真的吗？数组的大小将减少到10或者数组大小保持在80并且字符串填充数组位置，从数组[0]开始。
为什么不找出来？
#include< stdio.h>
int main（int argc，char ** argv）{
char array [80] ="我很短;
printf（"％lu \ n"，（unsigned long）sizeof（array））; / *最佳格式
c89 * /
返回0;
}
tedu写道：cj写道：有人可以帮我解决这个问题吗？我想我已经把它缩小到4个中的2个但不确定。
char数组[80] ="我很矮;
这是真的吗？数组的大小将减少到10或者数组大小保持在80并且字符串填充数组位置，从数组[0]开始。
为什么不找出来？
#include< stdio.h>
int main（int argc，char ** argv）{
char array [80] ="我很短;
printf（"％lu \ n，（unsigned long）sizeof（array））; / *最佳格式为
c89 * /
返回0;
}
学习_why_不是更好这是如此，而不只是
展示最终答案？
行
char数组[80 ] ="我很短;
保留80个字节供阵列使用。这些可以任何方式填写
可以想象。它还指定了字符''我很短''，然后是从数组[0]开始的0
。其余部分未使用，但可以在需要时使用。
C中的字符数组不是字符串。 C没有字符串
类型。按照惯例（由许多标准库
函数识别）" strings"只是由零
字符终止的字符数组。文字字符串（用双引号括起来的字符串）
只是方便地表达这个而不用实际输入
终止零 - 这是隐含的。
我肯定有更好的（更多的PC？）方式这样说，但我认为
它比仅仅展示后效更有帮助。
干杯
弗拉基米尔
-
我的电子邮件地址是真实的，我读了它。
cj写道：
有人可以帮我解决这个问题吗？我想我已经把它缩小到4个中的2个但不确定。
char数组[80] ="我很矮;
这是真的吗？阵列的大小将减少到10
No.
/ * BEGIN new.c * /
#include< stdio.h>
int main（无效）
{
char数组[80] ="我很简短;
printf（" sizeof array is％u \ n"，（unsigned）sizeof array）;
puts（array） ;
返回0;
}
/ * END new.c * /
或
数组大小保持在80并且字符串填充数组位置，
从数组[0]开始。
是的，其余的字符串后面的数组元素，
初始化为零。
-
pete
Can someone help me with this question? I think I''ve narrowed it down
to 2 out of the 4 but not sure.
char array [80] = " I am short";
Which would be true? The size of the array would be reduced to 10 or
the array size stays at 80 and the string fills array locations,
starting at array [0].
cj wrote:Can someone help me with this question? I think I''ve narrowed it down
to 2 out of the 4 but not sure.
char array [80] = " I am short";
Which would be true? The size of the array would be reduced to 10 or
the array size stays at 80 and the string fills array locations,
starting at array [0].
why not find out?
#include <stdio.h>
int main(int argc, char **argv) {
char array [80] = " I am short";
printf("%lu\n", (unsigned long)sizeof(array)); /* best format for
c89 */
return 0;
}
tedu wrote:cj wrote:Can someone help me with this question? I think I''ve narrowed it down
to 2 out of the 4 but not sure.
char array [80] = " I am short";
Which would be true? The size of the array would be reduced to 10 or
the array size stays at 80 and the string fills array locations,
starting at array [0].
why not find out?
#include <stdio.h>
int main(int argc, char **argv) {
char array [80] = " I am short";
printf("%lu\n", (unsigned long)sizeof(array)); /* best format for
c89 */
return 0;
}
Wouldn''t it be better to learn _why_ this is so, instead of just
demonstrating the final answer?
The line
char array[80] = " I am short";
reserves 80 bytes for use by the array. These can be filled in any way
imaginable. It also assigns the characters '' I am short'' followed by a 0
starting with array[0]. The rest is unused, but can be if need be.
Character arrays in C are not "strings". C does not have a "string"
type. By convention (as recognised by numerous standard library
functions) "strings" are just arrays of characters terminated by a zero
character. Literal strings (the ones surrounded by double quotes) are
just convenient way to express this without actually typing the
terminating zero -- that is implied.
I''m sure there are better (more PC?) ways of saying this, but I think
it''s still more helpful than just demonstrating the after-effects.
Cheers
Vladimir
--
My e-mail address is real, and I read it.
cj wrote:
Can someone help me with this question? I think I''ve narrowed it down
to 2 out of the 4 but not sure.
char array [80] = " I am short";
Which would be true? The size of the array would be reduced to 10
No.
/* BEGIN new.c */
#include <stdio.h>
int main(void)
{
char array [80] = " I am short";
printf("sizeof array is %u\n", (unsigned) sizeof array);
puts(array);
return 0;
}
/* END new.c */
or
the array size stays at 80 and the string fills array locations,
starting at array [0].
Yes, and the rest of the array elements following the string,
are initialized to zero.
--
pete
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