浮动到int型铸造？ [英] Float to int type casting?
问题描述
1 int main（）
2 {
3浮动a = 17.5;
4 printf（"％d \ n"，a）;
5 printf（"％d \ n"，*（int *）& a）;
6返回0;
7}
输出结果为：
0
1099694080
我无法理解为什么第一个printf打印0和另一个
，1099694080 ...
我认为与简单的类型转换有所不同。这是一个
未定义的行为吗？
请让我理解。
问候。
Hi,
1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);
5 printf("%d\n", *(int *)&a);
6 return 0;
7 }
The output is:
0
1099694080
I could not understand why the first printf prints 0 and the other
,1099694080...
I think there is sth different from simple type casting.Is it an
undefined behaviour?
Please let me understand.
Regards.
推荐答案
Enekajmer写道：
Enekajmer wrote:
1 int main（）
2 {
3浮动a = 17.5;
4 printf（"％d \ n"，a）;
这里你试图用转换说明符打印浮点值
for int。结果未定义。
5 printf（"％d \ n"，*（int *）& a）;
在这里，你重新解释了浮点数的表示。结果
是实现定义的。
6返回0;
7}
输出结果为：
0
1099694080
我无法理解为什么第一个printf打印0和另一个
，1099694080 ......
我认为有与简单类型铸造不同。这是一种不确定的行为吗？
Hi,
1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);
Here you''re trying to print a float value with a conversion specifier
for int. The results are undefined.
5 printf("%d\n", *(int *)&a);
Here, you''re reinterpreting the representation of the float. The results
are implementation defined.
6 return 0;
7 }
The output is:
0
1099694080
I could not understand why the first printf prints 0 and the other
,1099694080...
I think there is sth different from simple type casting.Is it an
undefined behaviour?
[这个词是'某事'']
见上文。
HTH，
- g
-
Artie Gold - 德克萨斯州奥斯汀
http://goldsays.blogspot.com
http://www.cafepress.com/goldsays
如果你没有什么可隐瞒的，你就不会尝试！
[the word is `something'']
See above.
HTH,
--ag
--
Artie Gold -- Austin, Texas
http://goldsays.blogspot.com
http://www.cafepress.com/goldsays
"If you have nothing to hide, you''re not trying!"
Artie Gold写道：
Artie Gold wrote:
Enekajmer写道：
Enekajmer wrote:
1 int main（）
2 {
3浮动a = 17.5;
4 printf（"％d \ n"，a）;
Hi,
1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);
在这里，您尝试使用转换说明符
为int打印浮点值。结果未定义。
Here you''re trying to print a float value with a conversion specifier
for int. The results are undefined.
5 printf（"％d \ n"，*（int *）& a）;
5 printf("%d\n", *(int *)&a);
在这里，你重新解释浮动的表示。结果
是实现定义的。
Here, you''re reinterpreting the representation of the float. The results
are implementation defined.
6返回0;
7}
输出结果为：
0
1099694080
我无法理解为什么第一个printf打印0和另一个
，1099694080 ......
我认为有......与简单类型铸造不同。这是一种未定义的行为吗？
6 return 0;
7 }
The output is:
0
1099694080
I could not understand why the first printf prints 0 and the other
,1099694080...
I think there is sth different from simple type casting.Is it an
undefined behaviour?
[这个词是'某事'']
见上文。
HTH，
- g
-
Artie Gold - 德克萨斯州奥斯汀
http://goldsays.blogspot.com
http://www.cafepress.com/goldsays
如果你没有什么可隐瞒的，你就不会尝试！
[the word is `something'']
See above.
HTH,
--ag
--
Artie Gold -- Austin, Texas
http://goldsays.blogspot.com
http://www.cafepress.com/goldsays
"If you have nothing to hide, you''re not trying!"
谢谢我了解了行为的类型，但我想了解更多细节的原因。举个例子，这个片段没有问题
以下代码：
....
浮动a = 17.5;
int b = a; （从float到int的类型转换）
printf（" b - >％d"，b）; //通常情况下，打印b - > 17
....
根据这个例子，我猜到了输出我编译之前的代码也是17
。相比之下，它的输出为0 ...
printf（a - >％d，a）// a是浮点型
那么，这种情况的原因是什么？IEEE-754？
问候。
Thanks i learnt the types of behaviours but i want to learn causes in
more details.To give an example, there is no problem for the snippet
code below:
....
float a=17.5;
int b=a; (type casting from float to int)
printf("b-->%d",b); // as normally, this prints "b-->17"
....
according to this example i guessed the output of the code below as 17
too before i compiled. As contrast its output is 0...
printf("a-->%d",a) //a is in float type
So, what are the causes of this case?The IEEE-754?
Regards.
" Artie Gold" < AR ******* @ austin.rr.com>在消息中写道
news：44 ************ @ individual.net ...
"Artie Gold" <ar*******@austin.rr.com> wrote in message
news:44************@individual.net...
Enekajmer写道：
Enekajmer wrote:
1 int main（）
2 {
3浮动a = 17.5;
4 printf（"％d \ n，a ）;
Hi,
1 int main()
2 {
3 float a = 17.5;
4 printf("%d\n", a);
在这里，您尝试使用转换说明符
为int打印浮点值。结果是未定义的。
Here you''re trying to print a float value with a conversion specifier
for int. The results are undefined.
实际上，你正在尝试打印一个double值，因为当它''时，浮动是
。不受原型中正式参数的约束。
如果不是这样的话，你可能会得到两个相同的输出
printfs。
-
RSH
Actually, you''re trying to print a double value, because the float is
promoted when it''s not governed by a formal parameter in a prototype.
If it weren''t for that, you''d probably get the same output from both
printfs.
--
RSH
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