产量 [英] output

问题描述

o / p这个什么都没有..你可以解释一下为什么......我知道它会与签名和未签名的东西有关...


#include< stdio.h>


#define TOTAL_ELEMENTS（sizeof（array）/ sizeof（array [0]））

int array [] = {23,34,12,17,204,99,16};


int main（）

{

int d;

for（d = -1; d< =（TOTAL_ELEMENTS-2）; d ++）

printf（"％d \ n"，array [d + 1]）;


返回0;

}

解决方案

Abhishek Parwal写道：

o / p这个什么都没有..你可以解释为什么......我知道它会与签名和未签名的事情有关。

＃include< stdio.h>

#define TOTAL_ELEMENTS（sizeof（array）/ sizeof（array [0]））
int array [] = {23,34 ，12,17,204,99,16};

int main（）
{
int d;
for（d = -1; d< =（ TOTAL_ELEMENTS-2）; d ++）
printf（"％d \ n"，数组[d + 1]）;

返回0;
}

请在发布前搜索档案，这个具体问题已经好几次讨论过
：

http://groups.google.com/group/comp....f96e862940333b
http://groups.google.com/group /comp....1a62b1532e620d
http://groups.google.com/group/comp....18e466cd0ae5c5

Robert Gamble

Abhishek Parwal写道：

o / p这是noth你可以解释为什么......我知道它会与签名和未签名有关...

＃include< stdio.h>

＃定义TOTAL_ELEMENTS（sizeof（array）/ sizeof（array [0]））
int array [] = {23,34,12,17,204,99,16};

int main（对于（d = -1; d< =（TOTAL_ELEMENTS-2）; d ++）<<<<<<<<< sizeof是type size_t，而不是int
for（d = -1; d< =（int）（TOTAL_ELEMENTS-2）; d ++）printf（"％d \ n"，array [d + 1] ）;

返回0;
}

所以这里，size_t似乎是无符号的，因此d被提升

未签名用于比较，-1是一个非常大的正数，

bit-wise。你也可以使用一点点缩进。此外，数组可以在main（）中定义

。

-

- Mark

不，它与unsigned或signed无关，因为int是由

编译器默认采用签名（这可能是

实现依赖或编译依赖）。

你可以通过在声明前加上signed关键字来检查这个。


问题在于你的#define语句代码是

表达式。因此，它无法在for循环中与

变量d进行比较，从而导致无循环处理。

o/p of this is nothing .. can u explain why.. i know its gonna
something to do with signed and unsigned..

#include<stdio.h>

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {23,34,12,17,204,99,16};

int main()
{
int d;
for(d=-1;d <=(TOTAL_ELEMENTS-2);d++)
printf("%d\n",array[d+1]);

return 0;
}

解决方案

Abhishek Parwal wrote:

o/p of this is nothing .. can u explain why.. i know its gonna
something to do with signed and unsigned..

#include<stdio.h>

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {23,34,12,17,204,99,16};

int main()
{
int d;
for(d=-1;d <=(TOTAL_ELEMENTS-2);d++)
printf("%d\n",array[d+1]);

return 0;
}

Please search the archives before posting, this specific question has
been discussed several time before:

http://groups.google.com/group/comp....f96e862940333b
http://groups.google.com/group/comp....1a62b1532e620d
http://groups.google.com/group/comp....18e466cd0ae5c5

Robert Gamble

Abhishek Parwal wrote:

o/p of this is nothing .. can u explain why.. i know its gonna
something to do with signed and unsigned..

#include<stdio.h>

#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {23,34,12,17,204,99,16};

int main()
{
int d;
for(d=-1;d <=(TOTAL_ELEMENTS-2);d++) <<< sizeof is type size_t, not int for(d=-1;d <=(int) (TOTAL_ELEMENTS-2);d++) printf("%d\n",array[d+1]);

return 0;
}

So here, size_t appears to be unsigned thus ''d'' gets promoted to
unsigned for the comparison and -1 is a very big positive number,
bit-wise. You could use a little indenting too. Also, array can be
defined within main().
--
- Mark

No, it has nothing to do with the unsigned or signed because int is by
default taken by the compiler as signed ( this however maybe
implementation dependent or compile dependent).
You can check this by prefixing the declaration with signed keyword.

The problem lies with the # define statement in you code which is an
expression. Thereby it cannot be compared in the for loop with the
variable d, resulting in no loop processing.

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