问候，这是我编程的第一年和第3周的这个 所以算我是一个纯粹的新手。 我正在尝试制作一个程序一个数字和打印开始 如下： 输入3： * ** *** 输入6： * ** *** **** ***** ****** 正如您所看到的那样，我尝试制作的程序会产生与输入数字相同的行数 ，每行增加一颗星。 这个是我写的程序： #include< stdio.h> star（int x） { int i = 1; while（i< = x） { printf（" ; *"）; i ++; } } print（int x ）{ int start = 1; int end = x; while（start< = end） { printf（"％d \ n"，sta r（开始））; 开始++; } } main（） { printf（输入值\ n）; int veri; scanf（"％d" ，& veri）; print（veri）; } 似乎有效，但有一个错误： 我总是在这样的星星后得到一个数字： Ex： 输入值 8 * 2 ** 3 *** 4 **** 5 ***** 6 ****** 7 ******* 8 ** ****** 9 提前致谢:)
Greetings , This is my first year at programming and 3rd week at this so count me as a pure newbie. I am trying to make a program that gets a number and prints out starts as this: For input 3 : * ** *** For input 6: * ** *** **** ***** ****** As you see the program I am trying to make generates as much as lines as the input number , one star increasement at each line. This is the program I wrote: #include <stdio.h> star(int x) { int i=1; while(i<=x) { printf("*"); i++; } } print(int x){ int start = 1; int end = x; while (start <= end) { printf("%d \n", star(start) ); start++; } } main() { printf("Enter value \n"); int veri; scanf("%d", &veri); print(veri); } Seems works but with a bug: I always get a number after stars such as this: Ex: Enter value 8 *2 **3 ***4 ****5 *****6 ******7 *******8 ********9 Thanks in advance :)
推荐答案Odinn写道： Odinn wrote: 问候，这是我编程的第一年和第3周的这个 所以算我是一个纯粹的新手。 /> 我正在尝试创建一个获得数字并打印出来的程序 如下： 输入3： * ** *** 输入6： * ** *** **** ***** ****** 正如您所看到的我正在尝试的程序生成尽可能多的行 作为输入数字，每行增加一个星。 这是我写的程序： #include< stdio.h> star（int x） { int i = 1 ; while（i< = x） { printf（" *"）; i ++; } } print（int x）{ int start = 1 ; int end = x; while（start< = end） { printf（" ;％d \ n"，star（start））; start ++; } } main（） { printf（" Enter value \ n"）; int veri; scanf（"％d"，& veri）; print（veri）; } 似乎有效，但有一个错误： 我总是在这样的星星后得到一个数字： Ex： 输入值 8 * 2 ** 3 *** 4 **** 5 ***** 6 ****** 7 ******* 8 ******** 9 提前致谢:) Greetings , This is my first year at programming and 3rd week at this so count me as a pure newbie. I am trying to make a program that gets a number and prints out starts as this: For input 3 : * ** *** For input 6: * ** *** **** ***** ****** As you see the program I am trying to make generates as much as lines as the input number , one star increasement at each line. This is the program I wrote: #include <stdio.h> star(int x) { int i=1; while(i<=x) { printf("*"); i++; } } print(int x){ int start = 1; int end = x; while (start <= end) { printf("%d \n", star(start) ); start++; } } main() { printf("Enter value \n"); int veri; scanf("%d", &veri); print(veri); } Seems works but with a bug: I always get a number after stars such as this: Ex: Enter value 8 *2 **3 ***4 ****5 *****6 ******7 *******8 ********9 Thanks in advance :)
不确定你想要完成什么通过在printf中调用star 函数。也许我错过了什么。如果你只需要把它拿出去并在你的明星 函数中添加新行的打印，它似乎输出你所描述的想要的东西： #include< stdio.h> star（int x） { int i = 1; while（i< = x） { printf（" *"）; i ++; } printf（" \ n"）; } printstar（int x）{ int start = 1; int end = x; while（start< = end） { star（start）; start ++; } } main（） { printf（"输入值\ n"）; int veri; scanf（"％d"，& veri）; printstar（veri）; }
Not sure what you were trying to accomplish by calling the star function from within the printf. Maybe I am missing something. If you just take that out and add the printing of the new line in your star function it seems to output what you described as wanting: #include <stdio.h> star(int x) { int i=1; while(i<=x) { printf("*"); i++; } printf("\n"); } printstar(int x){ int start = 1; int end = x; while (start <= end) { star(start) ; start++; } } main() { printf("Enter value \n"); int veri; scanf("%d", &veri); printstar(veri); }
Odinn< da ******* @ gmailwrote： Odinn <da*******@gmailwrote: 问候，这是我编程的第一年第3周这个 所以算我是纯新手。 Greetings , This is my first year at programming and 3rd week at this so count me as a pure newbie.
嗯，你至少尝试过，这是一个很好的开始。还不错 所有，但有一些事情需要注意：
Well, you''ve tried at least, and that''s a great start. Not bad at all, but there are some things to note:
#include< stdio.h> #include <stdio.h>star（int x） star(int x)
这是有效的 - 返回类型默认为" INT"如果你没有明确地提供 - 但这不是一个好主意，它也不是最新的C语言标准所支持的。鉴于你没有给b $ b返回任何东西，也许是无效。在这里是合适的。
This is valid - the return type defaults to "int" if you don''t supply one explicitly - but it isn''t a great idea, and it also isn''t supported by the latest C language standard. Given that you don''t return anything, perhaps "void" would be appropriate here.
{ int i = 1; while（i< = x） { printf（" *"）; i ++; } } { int i=1; while(i<=x) { printf("*"); i++; } }print（int x）{ int start = 1; int end = x; while（start< = end） { printf（"％d \ n"，明星（开始））; print(int x){ int start = 1; int end = x; while (start <= end) { printf("%d \n", star(start) );
这里是你看到的那些数字进来的地方。在printf中的％d（） 告诉它看起来对于一个数字，这是星（开始）返回的。 因为你实际上没有返回任何东西，所以 被打印的数字不能被预测。请注意这一点。 star（start）; printf（" \ n"）; $ b如果没有打印那个数字，$ b就可以很好地工作。
Here''s where those numbers you''re seeing come in. That %d in printf() tells it to look for a number, which is what star(start) returns. Since you didn''t actually return anything, though, the number that gets printed can''t be predicted. Watch out for that. star(start); printf( "\n" ); will work quite nicely without printing that number.
start ++; } } start++; } }main（） main()
这样会更好 int main（无效） ....
This would be better as int main(void) ....
{ printf （输入值\ n）; int veri; scanf（"％d"，& veri）; print（veri）; { printf("Enter value \n"); int veri; scanf("%d", &veri); print(veri);....这意味着你会想要在这里返回一些东西。 0表示成功， 所以 返回0;
....which means you''ll want to return something here. 0 means success, so return 0;
} }可能比我的第一个C程序好... - C. Benson Manica |我*应该*知道我在说什么 - 如果我 cbmanica（at）gmail |不，我需要知道。火焰欢迎。
Probably better than my first C program... -- C. Benson Manica | I *should* know what I''m talking about - if I cbmanica(at)gmail | don''t, I need to know. Flames welcome.
非常感谢克里斯托弗先生的帮助，以及帮助。真的 赞赏。 Uluc Aydin Christopher Benson-Manica yazdi： Thanks alot for the help Mr Christopher , aswell as the help. Really appreciated. Uluc Aydin Christopher Benson-Manica yazdi: Odinn< da ******* @ gmailwrote： Odinn <da*******@gmailwrote: 问候，这是我编程的第一年和第3周> 所以算我是一个纯粹的新手。 Greetings , This is my first year at programming and 3rd week at this so count me as a pure newbie.
嗯，你至少尝试过，这是一个很好的开始。还不错 所有，但有一些事情需要注意：
Well, you''ve tried at least, and that''s a great start. Not bad at all, but there are some things to note:
#include< stdio.h> #include <stdio.h>star（int x） star(int x)
这是有效的 - 返回类型默认为" INT"如果你没有明确地提供 - 但这不是一个好主意，它也不是最新的C语言标准所支持的。鉴于你没有给b $ b返回任何东西，也许是无效。在这里是合适的。
This is valid - the return type defaults to "int" if you don''t supply one explicitly - but it isn''t a great idea, and it also isn''t supported by the latest C language standard. Given that you don''t return anything, perhaps "void" would be appropriate here.
{ int i = 1; while（i< = x） { printf（" *"）; i ++; } } { int i=1; while(i<=x) { printf("*"); i++; } }print（int x）{ int start = 1; int end = x; while（start< = end） { printf（"％d \ n"，明星（开始））; print(int x){ int start = 1; int end = x; while (start <= end) { printf("%d \n", star(start) );
这里是你看到的那些数字进来的地方。在printf中的％d（） 告诉它看起来对于一个数字，这是星（开始）返回的。 因为你实际上没有返回任何东西，所以 被打印的数字不能被预测。请注意这一点。 star（start）; printf（" \ n"）; $ b如果没有打印那个数字，$ b就可以很好地工作。
Here''s where those numbers you''re seeing come in. That %d in printf() tells it to look for a number, which is what star(start) returns. Since you didn''t actually return anything, though, the number that gets printed can''t be predicted. Watch out for that. star(start); printf( "\n" ); will work quite nicely without printing that number.
start ++; } } start++; } }main（） main()
这样会更好 int main（无效） ...
This would be better as int main(void) ...
{ printf（" Enter value \ n"）; int veri; scanf（"％d"，& veri）; print（veri）; { printf("Enter value \n"); int veri; scanf("%d", &veri); print(veri);...这意味着你会想要在这里归还一些东西。 0表示成功， 所以 返回0;
...which means you''ll want to return something here. 0 means success, so return 0;
} }可能比我的第一个C程序好... - C. Benson Manica |我*应该*知道我在说什么 - 如果我 cbmanica（at）gmail |不，我需要知道。火焰欢迎。
Probably better than my first C program... -- C. Benson Manica | I *should* know what I''m talking about - if I cbmanica(at)gmail | don''t, I need to know. Flames welcome.
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一个简单的程序,需要看看哪里有错误:)
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