这个功能有什么回报！ [英] What does this function return!
问题描述
#define UWORD2 unsigned short int
typedef BYTE Boolean;
Boolean evaluateBit（UWORD2 * value，int bitnumber）;
Boolean evaluateBit （UWORD2 * value，int bitnumber）{
int one = 1;
return（one&（* value> bitnumber））;
}
当我这样调用函数时：
#include< stdlib.h>
#define UWORD2 unsigned short int
typedef BYTE Boolean;
Boolean evaluateBit（UWORD2 * value，int bitnumber）;
int main（void ）{
布尔岛;
int row = 15;
int col = 15;
int i， j;
float qual [row] [col];
/ *另一个函数将值赋给qual并同时施放
他们漂浮* /
/ *以下浮动值被重新编号为unsigned char，然后才发送给函数* /
for（i = 0; i< col; i ++）{
for（j = 0; j< row; j ++）{
island = evaluateBit （（unsigned char）& qual [i] [j]，4）;
}
}
return（1）
}
布尔值evaluateBit（UWORD2 *值，int位数）{
int one = 1;
return（1&（* value> bitnumber））;
}
我收到以下警告：
警告：传递`evaluateBit'的arg 1'使得整数指针
没有演员阵容
我不理解返回语句和发生此警告的原因。
提前致谢，
谢尔顿
"谢尔顿" < sh ****** @ gmail.comwrites：
请使用缩进。
#include< ;文件stdlib.h>
#include< limit.h>
#define UWORD2 unsigned short int
typedef BYTE布尔值;
BYTE未定义
Boolean evaluateBit（UWORD2 * value，int bitnumber）;
char evaluateBit（unsigned char * value，unsigned bitnumber）{
int main（void）{
布尔岛;
int row = 15，col = 15;
int i，j;
浮动qual [row] [col];
/ *另一个函数将值赋给qual并同时
将它们转换为浮动* /
$ b $
被发送到函数* /
for（i = 0; i< col; i ++）{<之前b / *浮点值以下被重新编号为unsigned char br />
for（j = 0; j< row; j ++）{
island = evaluateBit（（unsigned char）& qual [i] [j]，4）;
evaluateBit需要一个指向unsigned short的指针而不是
一个unsigned char值。你想要的是：
island = evaluateBit（（unsigned short *）& qual [i] [j]，4）;
但是仍然可能存在一些问题。您应该使用：
island = evaluateBit（（unsigned char *）& qual [i] [j]，4）;
并更改evaluateBit函数。
BTW。我希望你打算和岛上做点什么。
}
}
返回1
您确定要返回1吗？
}
Boolean evaluateBit（UWORD2 * value，int bitnumber）{
int one = 1;
return（one&（* value> bitnumber））;
}
试试这个：
char evaluateBit（unsigned char * value，unsigned bitnumber）{
return（值[bitnumber / CHAR_BIT]>（bitnumber％CHAR_BIT））& 1;
}
我收到以下警告：
"警告：传递arg 1的` evaluateBit''从整数制作指针
没有演员表'
我不理解return语句
哪个退货单？
以及出现此警告的原因。
我希望我已经解释过了。
-
祝你好运， _ _
.o。 | Serenly Enlightened Majesty of o''，=。/`o
..o |计算机科学，Michalmina86 Nazarewicz（oo）
ooo + - < mina86 * tlen.pl> ---< jid：mina86 * chrome.pl> - ooO - （_） - Ooo--
在文章< 11 ********************** @ b28g2000cwb.googlegroups .com>中，
Sheldon< sh ****** @ gmail.comwrote：
> island = evaluateBit（（ unsigned char）& qual [i] [j]，4）;
> Boolean evaluateBit（UWORD2 * value，int bitnumber）{
>我收到以下警告：
"警告：传递`evaluateBit'的arg 1'使得整数指针
没有强制转换
>我不理解return语句以及为什么会出现此警告。
注意岛屿=你在哪里调用evaluateBit的作业。
你拿& qual [i] [j]然后把它投到（unsigned char）。
unsigned char不是指针，但evaluateBit需要一个指针
作为它的第一个参数。
可能你想要演绎（unsigned char *）
-
那时候我还很年轻，但我也很朦胧。 br />
- Christopher Priest
Sheldon< sh ****** @ gmail.comwrote：
浮动值以下的
/ *被重新编号为unsigned char，然后发送给函数* /
for（i = 0; i< col; i ++）{
for（j = 0; j< row; j ++）{
island = evaluateBit（（unsigned char）& qual [i] [j ]，4）;
与您的评论所说的相反，您不会施放浮动值，
您投下地址。然后你把一个转换地址（它现在是一个unsigned int类型的
）传递给一个函数，该函数需要一个地址
作为它的第一个参数，正如编译器告诉的那样你用
"警告：传递`evaluateBit'的arg 1'使得整数指针
没有强制转换
因为你不能取表达式的地址（即
''（unsigned int）qual [i] [j]''你实际上似乎想要使用），
你需要一个额外的变量来存储演员的结果和
然后将它的地址传递给你的函数：
for（i = 0; i< col; i ++）{
for（j = 0; j< row; j ++）{
unsigned int val =（unsigned int）qual [i] [j];
island = evaluateBit（& val，4）;
}
}
当然，如果你只是改变你的
函数来取一个unsigned int就可以变得更简单了它的第一个参数是
的地址...
问候，Jens
-
\ Jens Thoms Toerring ___ jt@toerring.de
\ __________________________ http://toerring.de
#define UWORD2 unsigned short int
typedef BYTE Boolean;
Boolean evaluateBit(UWORD2 *value, int bitnumber);
Boolean evaluateBit(UWORD2 *value, int bitnumber) {
int one=1;
return (one &(*value >bitnumber));
}
When I call the function like this:
#include <stdlib.h>
#define UWORD2 unsigned short int
typedef BYTE Boolean;
Boolean evaluateBit(UWORD2 *value, int bitnumber);
int main(void) {
Boolean island;
int row=15;
int col=15;
int i,j;
float qual[row][col];
/* another function assigns values to qual and at the same time casts
them to float */
/* below the float values are recasted to unsigned char before being
sent to the function */
for (i=0;i<col;i++){
for (j=0;j<row;j++){
island = evaluateBit((unsigned char)&qual[i][j],4);
}
}
return (1)
}
Boolean evaluateBit(UWORD2 *value, int bitnumber) {
int one=1;
return (one &(*value >bitnumber));
}
I get the following warning:
"warning: passing arg 1 of `evaluateBit'' makes pointer from integer
without a cast"
I don''t understand the return statement and why this warning occurs.
Thanks in advance,
Sheldon
"Sheldon" <sh******@gmail.comwrites:
Please, use indention.
#include <stdlib.h>#include <limit.h>
#define UWORD2 unsigned short int
typedef BYTE Boolean;BYTE is undefined
Boolean evaluateBit(UWORD2 *value, int bitnumber);char evaluateBit(unsigned char *value, unsigned bitnumber) {
int main(void) {
Boolean island;
int row = 15, col = 15;
int i, j;
float qual[row][col];
/* another function assigns values to qual and at the same time
casts them to float */
/* below the float values are recasted to unsigned char before
being sent to the function */
for (i = 0; i<col; i++) {
for (j = 0; j<row; j++) {
island = evaluateBit((unsigned char)&qual[i][j],4);evaluateBit expects a pointer to unsigned short instead it gets
an unsigned char value. What you want is:
island = evaluateBit((unsigned short*)&qual[i][j], 4);
but still there may be some problems with that. You should rather use:
island = evaluateBit((unsigned char*)&qual[i][j], 4);
and change evaluateBit function.
BTW. I do hope you intend to do something with island.
}
}
return 1Are you sure you want to return 1?
}
Boolean evaluateBit(UWORD2 *value, int bitnumber) {
int one = 1;
return (one &(*value >bitnumber));
}Try this one:
char evaluateBit(unsigned char *value, unsigned bitnumber) {
return (value[bitnumber / CHAR_BIT] >(bitnumber % CHAR_BIT)) & 1;
}
I get the following warning:
"warning: passing arg 1 of `evaluateBit'' makes pointer from integer
without a cast"
I don''t understand the return statementWhich return statement?
and why this warning occurs.I hope I''ve explained that.
--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o'' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
ooo +--<mina86*tlen.pl>---<jid:mina86*chrome.pl>--ooO--(_)--Ooo--
In article <11**********************@b28g2000cwb.googlegroups .com>,
Sheldon <sh******@gmail.comwrote:
>island = evaluateBit((unsigned char)&qual[i][j],4);
>Boolean evaluateBit(UWORD2 *value, int bitnumber) {
>I get the following warning:
"warning: passing arg 1 of `evaluateBit'' makes pointer from integer
without a cast"
>I don''t understand the return statement and why this warning occurs.Notice the island = assignment where you call evaluateBit.
You take &qual[i][j] and you cast that to (unsigned char).
An unsigned char is not a pointer, but evaluateBit needs a pointer
for its first argument.
Possibly you wanted to cast to (unsigned char*)
--
I was very young in those days, but I was also rather dim.
-- Christopher Priest
Sheldon <sh******@gmail.comwrote:/* below the float values are recasted to unsigned char before being
sent to the function */
for (i=0;i<col;i++){
for (j=0;j<row;j++){
island = evaluateBit((unsigned char)&qual[i][j],4);Contrary to what your comment says you don''t cast the float value,
you cast its address. And then you pass the cast address (which has
now a type of unsigned int) to a function that expects an address
as its first argument, exactly as the compiler tells you with
"warning: passing arg 1 of `evaluateBit'' makes pointer from integer
without a cast"Since you can''t take the address of of an expression (i.e. the
''(unsigned int) qual[i][j]'' you actually seem to want to use),
you need an extra variable to store the result of the cast and
then pass its address to your function:
for (i = 0; i < col; i++) {
for (j = 0; j < row; j++) {
unsigned int val = (unsigned int) qual[i][j];
island = evaluateBit(&val, 4);
}
}
Of course, it could get simpler if you would just change your
function to take an unsigned int as its first argument instead
of an address...
Regards, Jens
--
\ Jens Thoms Toerring ___ jt@toerring.de
\__________________________ http://toerring.de
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