输出数组的大小 [英] Outputting the size of an array
问题描述
我需要以下代码来返回存储在
数组中的字符串，并返回数组的大小。我想在没有
的情况下使用指针进行操作，到目前为止，该字符存储工作正常，它只是我不能想到一种方式返回数组中
个字符的数量。有人帮帮我吗？
int inputPhrase（）
{
char msg [characters]，ch;
int i = 0;
int n = 5;
while（（ch = getchar（））！=''\ n ''）
{
msg [i ++] = ch;
}
msg [我] =''\'''; / *防止垃圾放在阵列的末尾
输出* /
i = 0;
而（msg [i]！=''\ 0''）
{
printf（"％c"，msg [i ++]）;
}
返回n;
}
int main（）
{
inputPhrase（）;
int size = inputPhrase（）;
printf（"％d \\ \\ n \\ n，尺寸）;
返回0;
}
I need the following code to return a string of character stored in
array and also return the size of the array. I want to do it without
using pointers and so far have the character storing working fine, it''s
just that I can''t really think of a way to return the number of
characters in the array. Anyone help me out?
int inputPhrase()
{
char msg[characters], ch;
int i = 0;
int n = 5;
while ((ch = getchar()) != ''\n'')
{
msg[i++] = ch;
}
msg[i] = ''\0''; /* prevents garbage from being place at end of array
output */
i = 0;
while (msg[i] != ''\0'')
{
printf("%c", msg[i++]);
}
return n;
}
int main()
{
inputPhrase();
int size = inputPhrase();
printf("%d\n", size);
return 0;
}
推荐答案
忽略n = 5;我只是用它来确保main是正确输出n，显然，n需要最初设置为0
然后增加到适当的字符数
数组。
Ignore the n = 5; I just used that to make sure that main was
outputting n correctly, obviously, n needs to be originally set to 0
then incremented up to the appropriate number of characters in the
array.
tigrfire写道：
tigrfire wrote:
我需要以下代码才能返回一个字符串存储在
数组中的字符，也返回数组的大小。我想在没有使用指针的情况下做到这一点，到目前为止，角色存储工作正常，它只是因为我不能想到一种方法来返回
数组中的字符。有人帮帮我吗？
int inputPhrase（）
{char msg [characters]，ch;
int i = 0;
int n = 5;
while（（ch = getchar（））！=''\ n''）
{
msg [i ++] = ch;
}
msg [i] =''\'''; / *防止垃圾放在阵列末端
输出* /
i = 0;
while（msg [i]！=''\ 0''）
{
printf（"％c"，msg [i ++]）;
}
返回n;
}
int main（）
{
inputPhrase（）;
int size = inputPhrase（）;
printf（"％d \ n"，size）;
返回0;
}
I need the following code to return a string of character stored in
array and also return the size of the array. I want to do it without
using pointers and so far have the character storing working fine, it''s
just that I can''t really think of a way to return the number of
characters in the array. Anyone help me out?
int inputPhrase()
{
char msg[characters], ch;
int i = 0;
int n = 5;
while ((ch = getchar()) != ''\n'')
{
msg[i++] = ch;
}
msg[i] = ''\0''; /* prevents garbage from being place at end of array
output */
i = 0;
while (msg[i] != ''\0'')
{
printf("%c", msg[i++]);
}
return n;
}
int main()
{
inputPhrase();
int size = inputPhrase();
printf("%d\n", size);
return 0;
}
你已初始化n，但从未使用过它！
while（（ch = getchar（））！=''\ n''）
{
msg [i ++] = ch;
}
n = i; //这应该设置n的值。
此外，
You have initialized n, but never used it !
while ((ch = getchar()) != ''\n'')
{
msg[i++] = ch;
}
n = i; // This should set the value of n .
Further ,
char msg [characters]，
char msg[characters],
这不是一个有效的语句，无法使用
变量初始化数组。
This is not a valid statement, an array cannot be initialized with a
variable.
在函数外部我将字符定义为：
#define characters 30
如果我按照你说的去做并修改我的代码以便n =我在线上你
参考，程序的输出是这样的：
这是样本输出//输入的字符串
这是样本输出//返回字符串
0 //值n
Outside the function I defined characters as:
#define characters 30
If I do as you say and modify my code so that n = i on the line you
reference to, the program''s output is this:
this is sample output //entered string of chars
this is sample output //returned string of chars
0 //the value of n
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