atoi转换错了？

据我所知，这不是ATOI工作的方式。首先，您需要为NUBLIN输入数组中的空终止符留出空间。接下来，试试这样的事情：



      以下为原文

    To my understanding, that is not how atoi works. First, you'll need space for the null terminator in your numberInput array. Next, try something like this:
 

char numberInput[11];
numberInput[0] = '4';
numberInput[1] = '2';
numberInput[2] = '9';
numberInput[3] = '4';
numberInput[4] = '9';
numberInput[5] = '3';
numberInput[6] = '3';
numberInput[7] =' 6';
numberInput[8] = '9';
numberInput[9] = '5';
numberInput[10] = '';

首先，你没有终止你的字符串。最后你需要一个零。而且输入数需要允许长度为零。它可能只是工作，因为你碰巧跟随另一个变量，一开始就被初始化为零。ATO（）返回INT而不是未签名的long。使用正确的工作功能！



      以下为原文

    First you have not terminated your string.  You need a zero on the end.  And also a InputNumber needs to allow for that zero in the length. Its is probably only just working because you happen to follow that variable with another that happens to by initialized to zero at the start!
 
atoi() returns an int not an unsigned long.  Use the correct function for the job!

INTX= ATOI（“295”）；//ASCII到INTEGHYLY＝ATOL（“4294933695”）；//ASCII到龙龙Z= ATOL（“4294933695”）；PrINTF（“签名%LIR”，Z）；PrINTF（“未签名的%LUR”，Z）；



      以下为原文

     
int x = atoi("295");                 //ascii to integer
long y = atol("4294933695");    //ascii to long
 
long z = atol("4294933695");
printf("signed %lir",z);
printf("unsigned %lur",z);

首先，我认为OP有一个空终止的字符串，只是在他的帖子中重新键入错误；否则，他不会得到他发布的结果。@ OP，你的结果的原因是因为你使用了错误的转换函数。ATOI（）函数将字符串转换为16位整数。4294933695的下16位是0x7CBF，符号在分配给未签名的long之前扩展到0x000。7CBF，而4294950495的16位是0xBE5F，符号扩展到0xFFFFEB5F。对于您的输入号，需要使用将字符串转换成32位长整数的AtOLL（）函数。



      以下为原文

   
First, I think OP does have a null-terminated string and just retyped wrong in his post; otherwise, he would not get the results that he has posted.
 
@OP, the reason for your results is because you used the wrong conversion function. The atoi() function converts a character string to a 16-bit integer. The lower 16-bit of 4294933695 is 0x7CBF and sign-extends to 0x00007CBF before assigning to the unsigned long, while the lower 16-bit of 4294950495 is 0xBE5F and sign-extends to 0xFFFFBE5F.
 
For your input numbers, you need to use the atol() function that converts a string to a 32-bit long integer.

真滑稽——看（和评估）我的签名！！！！多个错误可以弥补一个（只是看起来）正确的结果。忽略丢失的ASCIZ终止，集中在数学上：ATOI（）返回一个整数0x8000。0x7FFF。Soatoi（4294933695）＝0x7CBATOOI（4294950495）＝0xBE5FFNE您肯定PrTrF（“…”%L……“”的值。这是怎么回事？”在“Prtff（）”参数中看到“%L”，编译器将相应的参数从短INT（INT16YT）传播到long int（INT32×T）。这个扩展的规则是将因特16t的位15（MSBIT）复制到因特32位（临时）变量的高位字的所有位。7CBF已清除位15 -它传播到0000 7CBFBE5F，位15集-传播到FFFF BE5FHT，看起来好像转换为RI。GHT为4294950495，但它从来没有。它不是编译器或图书馆的故障-它只是PEBKAC。



      以下为原文

    Really funny - see (and evaluate) my signature !!!
Multiple faults can make up for a (only seemingly) correct result.
 
Ignoring the missing ASCIIZ termination, concentrating on the maths:
atoi() returns an integer 0x8000 .. 0x7FFF. So
atoi(4294933695) = 0x7CBF
atoi(4294950495) = 0xBE5F
 
Now you certainly printf("... %l ...")'ed the values. What's happening with that?
"Seeing" the %l in the printf() argument, the compiler propagates the corresponding parameter from short int (int16_t) to long int (int32_t). And the rule for this expansion is to replicate Bit 15 (the MSBit) of the int16_t into all bits of the high-word of the int32_t (temporary) variable.
7CBF has bit 15 cleared  - this it propagates to 0000 7CBF
BE5F has bit 15 set - propagating to FFFF BE5F
 
Thus it looked as if the conversion had been right for 4294950495, but it never was.
 
It's not a compiler or library fault - it's just PEBKAC.

我曾经和一位年长的工程师（退休后）一起工作，他总是把这件事发送出去。



      以下为原文

   


I used to work with an older engineer (since retired) who was always sending this out :-).

谢谢大家的回复！我在原来的职位上犯了一个错误。编号输入〔0〕＝‘4’；不是4，与其余部分相同。这是原来用来输入较小数目的。我认为它能适用于长的数字，因为长的整数值就更大了。



      以下为原文

    Thank you all for the replies!
 
I made a mistake in my original post.    numberInput[0] = '4'; not 4 and same with the rest.
 
This was original used to input smaller numbers.  I thought it would work for long numbers because long is an integer value just bigger. 

“更大”正是关键所在：



      以下为原文

    "just bigger" is exactly the point  :)

如上所述，在C中工作时，必须始终知道变量的大小。在XC8中，一个“整数”变量总是16位。一个“长整数”（或者仅仅是“长”）总是32位。使用比你所需要的大的变量是浪费的，但是你必须始终确保它足够大以容纳NU。你正在工作。



      以下为原文

   
As above.
You must ALWAYS be aware of the size of your variables when working in C.
In XC8, an "integer" variable is always 16 bits.
A "long integer" (or just "long") is always 32 bits.
It's wasteful to use variables bigger than you need, but you must always make sure they are big enough to hold the numbers you are working with.
 

字符串实际上是由空字符“0”终止的一维字符数组。字符串中的第十一个字符位于数组外。是否使用调试器？我想它应该给你线索，但我是新来的；- Rob



      以下为原文

    Strings are actually one-dimensional array of characters terminated by a null character ''.
The eleventh character in your string is outside the array.
 
Are you using a debugger? I would think it should have given you a clue, but I'm new here ;-)
 
Rob