需要帮助找到为什么一个for循环的计数器变量被循环内的一个函数改变 [英] need help finding why a for loop's counter variable is being altered by a function inside the loop
问题描述
在我的循环内部的一个函数是以某种方式改变我正在迭代的值，我不知道如何。对不起，如果这是很不好描述。
里面这个for循环
int k; （k = 0; k <512; k ++）
{
// Discardheader（d）;
//实际上并没有做任何事情，因为它是一个header.f
int databit = Getexpecteddata（d + 4 * k + 1）;
printf（％d，k）;
int transmit =数据采样（& datastate，& datalength，d + 4 * k + 2，dataerr，dataloc，databit）;
printf（％d，k）;
Clocksample（& clockstate，& clocklength，d + 4 * k + 3，clockerr，transmit）;
printf（％d \\\
，k）;
$我得到这个输出
16 16 16
17 17 17
18 18 18
19 19 19
20 20 20
21 1 1
2 2 2
3 3 3
4 4 4
所以Datasample一旦达到21就改变k的值。d是char * d类型，表示一个缓冲区，在这里我读取一个文件。改变输入文件不会改变，在21开关发生。这里是datasample的代码：
int Datasample（int * state，int * length，char * d，int * type， int位置，int数据）
{
int match = 1;如果数据采样和延迟tx匹配，则
if（（（d [0]>> location）& 1）！= data）
{
match = 0;
if（data）
{
type [2] ++;
}
else
{
type [1] ++;
}
}
int ia; （（（d [0]>> ia）& 1）== *状态下的（bia = 7; ia> = 0; ia-）
（
） ）//找到一个边
{
* length ++;
}
else
{
int距离，偏差，devflag = 1; //边距离采样点的距离。应该是大约4
如果（位置> 3）//偏差距离那么距离是多远4
{distance = location-ia;}
else
{distance = ia - location;}
deviation = abs（4-distance）; （（偏差> = devmax）&& match&&& devflag）
{
devflag = 0;
if（data）
{
type [2] ++;
}
else
{
type [1] ++;
}
* state =（（d [0]> ia）& 1）;
* length = 1;
}
}
return（（d [0]>> location）& 1）;
}
一旦击中21，回滚到1？
在此先感谢。我不知道我在做什么。
解决方案我不知道程序应该做什么（注释并没有太大帮助，变量名称不是描述性的）。主要的问题似乎是 * lenght ++; 语句，这会使指针碰撞超出识别范围。接下来的 * length = 1; 是不好的工作。
额外的评论风格：对 unsigned 类型进行一些位操作;签名延伸可能会导致1位出现在不需要的地方。另外：建议使用无符号类型的计数器和索引;这会导致程序在下溢时严重崩溃。
$ $ $ $ $ $ $ $ $ $ $ $ int $ Datasample（int * state，int * length，char * d ，int *类型，无符号位置，int数据）
{
int match = 1; //如果数据采样和延迟tx匹配，
int devflag = 1; / *从内部循环中提升这个变量* /
unsigned bitpos; / *重命名并更改为无符号（位置以及）* /
/ *注意：移位为零（或负数）未定义* /
if（（（d [0]> gt ;>位置）& 1）！= data）{
match = 0;
if（data）type [2] ++;
else type [1] ++;
再次：零位移位未定义* /
（bitpos = 8; bitpos--> 0;）
{
// find （（（d [0]> bitpos）& 1）== *状态）*长度+ = 1的边缘
;
else
{
int距离，偏差;
//边缘距采样点的距离。应该是大约4
//偏差距离距离4
distance =（location> 3）有多远？位置 - bitpos：bitpos - 位置;
deviation = abs（4-distance）;
if（deviation> = devmax&& match&& devflag）
{
devflag = 0;
if（data）type [2] ++;
else type [1] ++;
}
* state =（（d [0]>> bitpos）& 1）;
* length = 1;
}
}
return（（d [0]>> location）& 1）;
$ b BTW是0_patterns的预期输出吗？
文件包含5769个事件
文件包含6938个错误
文件包含543个溢出错误
文件包含6395个非填充错误
错误nonspill＃spill＃
类型D 2250 451
类型C 0 0
类型B 4145 92
案例1 1195 307
案例2 0 20
案例3 1055 124
案例4 0 0
案例5 0 0
案例6 0 0
案例7 0 0
案例8 1160 9
案例9 0 0
案例10a 1472 39
案例10b 1513 29
案例10c 0 15
a function inside my loop is somehow changing the value that i am iterating over, and i'm not sure how. i'm sorry if this is very poorly described.
inside this for loop
int k;
for( k = 0; k < 512; k++)
{
// Discardheader(d); // doesnt actually do anything, since it's a header.f
int databit = Getexpecteddata(d+4*k+1);
printf("%d ",k);
int transmitted = Datasample(&datastate, &datalength, d+4*k+2,dataerr,dataloc, databit);
printf("%d ",k);
Clocksample(&clockstate, &clocklength, d+4*k+3,clockerr, transmitted);
printf("%d \n",k);
}
i get this output
16 16 16
17 17 17
18 18 18
19 19 19
20 20 20
21 1 1
2 2 2
3 3 3
4 4 4
so somehow Datasample is changing the value of k once it reaches 21. d is type char * d and represents a buffer where i read a file in. changing input files does not change that at 21 the switch happens. here is the code for datasample:
int Datasample (int* state, int* length, char *d, int *type, int location, int data)
{
int match = 1; // if data sample and delayed tx match,
if ( ((d[0] >> location) & 1) != data)
{
match = 0;
if(data)
{
type[2]++;
}
else
{
type[1]++;
}
}
int ia;
for( ia = 7; ia>=0; ia--)
{
if ( ((d[0] >> ia) & 1) == *state) // finds an edge
{
*length++;
}
else
{
int distance, deviation,devflag=1; // distance the edge is from the sample point. should be about 4
if ( location > 3) // deviation is how far the distance then is from 4
{distance = location - ia;}
else
{distance = ia - location;}
deviation = abs(4-distance);
if( (deviation >= devmax) && match && devflag)
{
devflag =0;
if(data)
{
type[2]++;
}
else
{
type[1]++;
}
}
*state = ((d[0] >> ia) & 1);
*length = 1;
}
}
return ((d[0] >> location) & 1);
}
what is causing the k to roll back to 1 once it hits 21?
thanks in advance. i have no idea what i'm doing.
解决方案 I don't have the faintest idea what the program is supposed to do (comments don't help very much, the variable names are not that descriptive). The main problem appears to be the *lenght++; statement, which bumps the pointer beyond recognition. A subsequent *length = 1; does the dirty work.
Extra comment on style: it is preferable to perform bit operations on unsigned types; sign-extention could cause '1' bits to appear at unwanted places. Also: it is advisable to use unsigned types for counters and indexes; that will cause the program to crash more rigorously on underflow.
int Datasample (int *state, int *length, char *d, int *type, unsigned location, int data)
{
int match = 1; // if data sample and delayed tx match,
int devflag = 1; /* hoisted this variable from inner loop */
unsigned bitpos ; /* renamed and changed to unsigned ( location as well) */
/* Note: shift by zero (or negative) is undefined */
if ( ((d[0] >> location) & 1) != data) {
match = 0;
if(data) type[2]++;
else type[1]++;
}
/* Again: shift by zero is undefined */
for( bitpos = 8; bitpos-- > 0; )
{
// find an edge
if ( ((d[0] >> bitpos) & 1) == *state) *length += 1;
else
{
int distance, deviation;
// distance the edge is from the sample point. should be about 4
// deviation is how far the distance then is from 4
distance = (location > 3) ? location - bitpos : bitpos - location;
deviation = abs(4-distance);
if (deviation >= devmax && match && devflag)
{
devflag =0;
if (data) type[2]++;
else type[1]++;
}
*state = ((d[0] >> bitpos) & 1);
*length = 1;
}
}
return ((d[0] >> location) & 1);
}
BTW is this the expected output for 0_patterns ?
File contains 5769 events
File contains 6938 errors
File contains 543 spill errors
File contains 6395 nonspill errors
Error nonspill # spill #
Type D 2250 451
Type C 0 0
Type B 4145 92
Case 1 1195 307
Case 2 0 20
Case 3 1055 124
Case 4 0 0
Case 5 0 0
Case 6 0 0
Case 7 0 0
Case 8 1160 9
Case 9 0 0
Case 10a 1472 39
Case 10b 1513 29
Case 10c 0 15
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