如何选择何时在字符数组中打印特定字符 [英] How do I select when to print specific characters in a character array
问题描述
对于我的作业,我需要执行以下操作:
For my assignment I need to do the following
- 请求用户输入他们的姓名并将其保存为字符数组
-
- 然后我需要输出用户输入的字符数。
- 然后我必须向后拼写用户名。
- 然后告诉用户字符数组的哪个部分包含他们的姓和名之间的空格
-
我需要做的最后一件事,但不能安静(实际上我完全失去了)输出用户的姓氏首先,然后名字最后(即John Doe = Doe John);
- Ask the user to input their name and save it to a character array
- Then I need to output the number of characters input by said user
- Then I have to spell that user's name backwards
- followed by telling the user which part of the character array contained the space between their first and last name
The last thing that I need to do but can't quiet figure out (actually I'm completely lost) is output the user's last name first then first name last (ie John Doe = Doe John);
#include <stdio.h>
#include <string.h>
int main ()
{
printf("Enter your name\n");
char name [25]={'\0'};
char * pch;
fgets(name,sizeof(name),stdin);
printf("You Entered: %s \n", name);
printf("There are %u characters in your name including the space. \n", strlen(name));
char end;
int i;
end = strlen(name) -1;
printf("Your name backwards is");
for (i = end; i >= 0; --i)
{
printf("%c", name [i]);
}
printf("\nLooking for the space in your name \n", name);
pch=strchr(name, ' ');
while (pch!=NULL)
{
printf("Space found at %d\n", pch-name+1);
pch=strchr(pch+1, ' ');
}
}
任何知道如何做的人你能解释你做了什么,所以我可以向你学习吗?
Anyone that knows how to do this Can you please explain what you did so I can learn from you?
推荐答案
我需要做的最后一件事, b $ b我完全丢失)输出用户姓氏首先然后首先
名称最后(即John Doe = Doe John);
The last thing that I need to do but can't quiet figure out (actually I'm completely lost) is output the users last name first then first name last (ie John Doe = Doe John);
这是一个常见的任务/面试问题,如何颠倒数组中的字符串。一个优雅的虽然可能不是最有效的算法是:
It is common assignment/interview question of how to reverse an oder of strings in an array. An elegant albeit maybe not the most efficient algorithm would be:
step 1: reverse the array character by character
step 2: reverse each string in-place
检查答案此处了解更多信息。
Check the answers here for more info.
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