gcc：error：case标签不会减少为整数常量 [英] gcc: error: case label does not reduce to an integer constant
问题描述
嗨再次，
这是我的问题：当我尝试编译下面包含
函数的代码时，我明白了：
-
gcc：21：错误：case标签没有减少到整数常量
gcc：24：错误：case标签不会减少到整数常数
-
现在，我（我认为我）理解错误，但是cryptMsg [i]肯定是
对我来说不是一个常数...
感谢您的帮助，
SysSpider
加密系统，虽然这对问题并不重要......
- 源代码 -
void decryptMsg（int sizeMsg）{
int i;
char cryptMsg [sizeMsg]，key [sizeMsg];
printf（&\\;输入加密的消息（无空格）："）;
scanf（ "％s"，cryptMsg）;
printf（"输入用过的密钥（无空格）："）;
scanf（"％s"，key）;
printf（" \ n消息是：\ n"）;
for（i = 0;我< sizeMsg; i ++）
{
if（i％2 == 0&& i！= 0）printf（""）;
switch（cryptMsg [i]）//这里开始问题
{
case charToInt（cryptMsg [i]）< charToInt（key [i]）：//错误
printf（"％d"，charToInt（cryptMsg [i]）+ 10 -
charToInt（key [i ]））;
break;
case charToInt（cryptMsg [i]）> = charToInt（key [i]）：//错误
printf（"％d"，charToInt（cryptMsg [i]） -
charToInt（key [i]））;
break;
默认：
printf（" \ nError！\ n"）;
休息;
}
}
}
- 来源结束 -
Hi again,
This is my problem: when i try to compile the code that contains the
function below, i get this:
--
gcc:21: error: case label does not reduce to an integer constant
gcc:24: error: case label does not reduce to an integer constant
--
Now, I (think that I) understand the error, but cryptMsg[i] surely
doesn''t look like a constant to me...
Thanks for help,
SysSpider
P.S.: btw, for those interested, I''m implementing the "one time pad"
cryptography system, although that isn''t important to the problem...
--Source code--
void decryptMsg(int sizeMsg) {
int i;
char cryptMsg[sizeMsg], key[sizeMsg];
printf("\nEnter the encrypted message (no spaces): ");
scanf("%s", cryptMsg);
printf("Enter the used key (no spaces): ");
scanf("%s", key);
printf("\nThe message is:\n");
for(i = 0; i < sizeMsg; i++)
{
if(i%2 == 0 && i != 0) printf(" ");
switch(cryptMsg[i]) //Here starts the problem
{
case charToInt(cryptMsg[i]) < charToInt(key[i]): //Error
printf("%d", charToInt(cryptMsg[i]) + 10 -
charToInt(key[i]));
break;
case charToInt(cryptMsg[i]) >= charToInt(key[i]): //Error
printf("%d", charToInt(cryptMsg[i]) -
charToInt(key[i]));
break;
default:
printf("\nError!\n");
break;
}
}
}
--End of source--
推荐答案
Sy*************@hackernetwork.com （ SysSpider）写道：
Sy*************@hackernetwork.com (SysSpider) writes:
-
gcc：21：错误：case标签没有缩小为整数常量
gcc：24：error：case label does不减少到一个整数常数
-
大概是这条线和它下面类似的线：
case charToInt（cryptMsg [i] ）< charToInt（key [i]）：//错误
现在，我（认为我）理解错误，但cryptMsg [i]肯定不会看起来像一个常数我...
--
gcc:21: error: case label does not reduce to an integer constant
gcc:24: error: case label does not reduce to an integer constant
--
Presumably for this line and the similar one below it:
case charToInt(cryptMsg[i]) < charToInt(key[i]): //Error
Now, I (think that I) understand the error, but cryptMsg[i] surely
doesn''t look like a constant to me...
完全正确。案例标签必须是整数常量。你不能用b / b
用switch语句做这种测试。我建议使用
if语句。
-
如果我告诉过你一次，我已经告诉过了你LLONG_MAX次不要夸大。
- 杰克克莱恩
Exactly. A case label must be an integer constant. You cannot
do this sort of test with a switch statement. I suggest using an
if statement instead.
--
"If I''ve told you once, I''ve told you LLONG_MAX times not to
exaggerate."
--Jack Klein
SysSpider写道：
SysSpider wrote:
嗨再次，
这是我的问题：当我尝试编译包含以下
函数的代码时，我得到了这个：
-
gcc：21：错误：case标签没有减少到整数常量
gcc：24：错误：case标签没有减少到整数常量
-
现在，我（认为我）理解错误，但cryptMsg [i]肯定对我来说看起来不是一个常数...
感谢您的帮助，SysSpider
PS：顺便说一句，对于那些感兴趣的人，我正在实施一次性密码加密系统，尽管这对于问题...
- 源代码 -
void decryptMsg（int sizeMsg）{
int i;
char cryptMsg [sizeMsg]，key [sizeMsg];
printf（" ; \ n输入加密的消息（无空格）："）;
scanf（"％s"，cryptMsg）;
printf（"输入用过的密钥（无空格）：" ）;
scanf（"％s"，key）;
printf（" \\\
The message is：\ n"）;
for （i = 0;我< sizeMsg; i ++）
{
if（i％2 == 0&& i！= 0）printf（""）;
switch（cryptMsg [i] ）//这里开始问题
{char / case charToInt（cryptMsg [i]）< charToInt（key [i]）：//错误
printf（"％d"，charToInt（cryptMsg [i]）+ 10 -
charToInt（key [i]））;
break;
case charToInt（cryptMsg [i]）> = charToInt（key [i]）：//错误
printf（"％d"，charToInt（cryptMsg [i]） -
charToInt（key [i]））;
break;
默认：
printf（" \ nError！\ n"）;
break;
}
}
}
- 来源结束 -
Hi again,
This is my problem: when i try to compile the code that contains the
function below, i get this:
--
gcc:21: error: case label does not reduce to an integer constant
gcc:24: error: case label does not reduce to an integer constant
--
Now, I (think that I) understand the error, but cryptMsg[i] surely
doesn''t look like a constant to me...
Thanks for help,
SysSpider
P.S.: btw, for those interested, I''m implementing the "one time pad"
cryptography system, although that isn''t important to the problem...
--Source code--
void decryptMsg(int sizeMsg) {
int i;
char cryptMsg[sizeMsg], key[sizeMsg];
printf("\nEnter the encrypted message (no spaces): ");
scanf("%s", cryptMsg);
printf("Enter the used key (no spaces): ");
scanf("%s", key);
printf("\nThe message is:\n");
for(i = 0; i < sizeMsg; i++)
{
if(i%2 == 0 && i != 0) printf(" ");
switch(cryptMsg[i]) //Here starts the problem
{
case charToInt(cryptMsg[i]) < charToInt(key[i]): //Error
printf("%d", charToInt(cryptMsg[i]) + 10 -
charToInt(key[i]));
break;
case charToInt(cryptMsg[i]) >= charToInt(key[i]): //Error
printf("%d", charToInt(cryptMsg[i]) -
charToInt(key[i]));
break;
default:
printf("\nError!\n");
break;
}
}
}
--End of source--
它不会像蜘蛛一样工作，给它与switch（）。 C'的情况
没有表达。它需要一个常数。你需要提前知道
的案例。
开关（getnum（））{
案例1 ：有趣（1）;休息;
案例2：有趣（2）;休息;
默认：有趣（3）;
}
规范的独家案例构造可以完成喜欢这个..
if（case1）
fun（1）;
else if（case2）
fun（2）;
else
fun（3）;
...其中case1和case2表达式为0或不为0。
-
Joe Wright mailto：jo ******** @ comcast.net
所有东西都应尽可能简单，但不能简单。
---阿尔伯特爱因斯坦---
It won''t work your way Spider, give it up with switch(). C''s case
does not take an expression. It takes a constant. You have to know
in advance what the cases are looking for.
switch(getnum()) {
case 1: fun(1); break;
case 2: fun(2); break;
default: fun(3);
}
The canonical ''exclusive case'' construct can be done like this..
if (case1)
fun(1);
else if (case2)
fun(2);
else
fun(3);
...where case1 and case2 are expressions of 0 or not-0.
--
Joe Wright mailto:jo********@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
Ben Pfaff< bl*@cs.stanford.edu>这样说：
Ben Pfaff <bl*@cs.stanford.edu> spoke thus:
案例标签必须是整数常量。
A case label must be an integer constant.
这与说案件相同标签必须有整数值
在编译时可计算，不是吗？ （fmi）
-
Christopher Benson-Manica |我*应该*知道我在说什么 - 如果我
ataru（at）cyberspace.org |不，我需要知道。火焰欢迎。
That''s the same as saying that case labels must have integer values
computable at compile time, isn''t it? (fmi)
--
Christopher Benson-Manica | I *should* know what I''m talking about - if I
ataru(at)cyberspace.org | don''t, I need to know. Flames welcome.
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