如果我连接到Heroku worker dyno heroku run(例如,heroku run python用于交互式Python会话),任何通过此方式显示Unicode字符的尝试都会导致UnicodeEncodeError
本地:
$ python
Python 2.7.1 (r271:86832, Jun 25 2011, 05:09:01)
[GCC 4.2.1 (Based on Apple Inc. build 5658) (LLVM build 2335.15.00)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> print u'\xa3'
£
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通过heroku run:
$ heroku run python
Running python attached to terminal... up, run.1
Python 2.7.2 (default, Oct 31 2011, 16:22:04)
[GCC 4.4.3] on linux2
Type "help", "copyright", "credits" or "license" for …Run Code Online (Sandbox Code Playgroud) 我有一个EXC_BAD_ACCESS在main(),这里是我的代码:
int main(int argc, char *argv[])
{
NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init];
int retVal = UIApplicationMain(argc, argv, nil, @"TestBedAppDelegate");
[pool release];
return retVal;
}
@interface TestBedAppDelegate : NSObject <UIApplicationDelegate>
@end
@implementation TestBedAppDelegate
- (void)applicationDidFinishLaunching:(UIApplication *)application {
UIWindow *window = [[UIWindow alloc] initWithFrame:[[UIScreen mainScreen] bounds]];
UINavigationController *nav = [[UINavigationController alloc] initWithRootViewController:[[TestBedViewController alloc] init]];
[window addSubview:nav.view];
[window makeKeyAndVisible];
}
@end
- (void) action: (id) sender
{
[self highRetainCount];
}
@implementation TestBedViewController
- (void) highRetainCount
{
UIView …Run Code Online (Sandbox Code Playgroud) 我正试图将一些C移植到Go.
我基本上在寻找类似于Haskell的东西 find :: (a -> Bool) -> [a] -> Maybe a
我(大致)这个C用于通过迭代查找"列表"中的项目:
struct foo {
struct foo *next;
char *name;
}
struct foo *foo_list;
// Snip
struct foo *foo = NULL;
for (f = foo_list; f; f = f->next) {
if (!strcmp("bar", f->name) {
foo = f;
}
}
if (foo)
// Stuff
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我怎么能在Go中很好地和惯用地做到这一点?
"名单"可能很小; 性能特征不是特别有趣.
我可能想要一个slice或一个list?Foos或*Foos的"列表" ?
我目前有以下内容,但我怀疑它不是特别"惯用Go"!
var FooTable *list.List
// Snip
var foo *Foo = nil
for e …Run Code Online (Sandbox Code Playgroud) 我有这个带有 ?: 运算符的表达式:
(adc.Voltage1[Counter.ADC_ConversionCount] = ADC_readResult(Handler.myAdc, ADC_ResultNumber_1)) > 10 ? Counter.WriteOut = 1 : Counter.WriteOut = 0;
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与 if-else 相同的表达式:
if((adc.Voltage1[Counter.ADC_ConversionCount] = ADC_readResult(Handler.myAdc, ADC_ResultNumber_1)) > 10 ){
Counter.WriteOut = 1;
}else{
Counter.WriteOut = 0;
}
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为什么在第一种情况下会出现“表达式必须是可修改的左值”错误?
ADC_readResult 函数返回类型是 uint_least16_t。这是 Counter 结构体定义和 ADC 结构体定义:
typedef struct __COUNTERS__ {
uint16_t WriteOut;
uint16_t ADC_ConversionCount;
uint16_t ADC_CycleCount;
uint8_t LimitADCReached1;
uint8_t LimitADCReached2;
uint8_t LimitADCReached3;
uint8_t LimitADCReached4;
uint8_t LimitADCReached5;
} COUNTERS;
typedef struct __ADC_VOLTAGES__ {
uint16_t Voltage1[ADC_VAL];
uint16_t Voltage2[ADC_VAL];
uint16_t Voltage3[ADC_VAL];
uint16_t Voltage4[ADC_VAL];
uint16_t Voltage5[ADC_VAL];
} …Run Code Online (Sandbox Code Playgroud) 我想根据某个变量以不同的方式打印同一页面的名称.
这是相应的代码.
$metaTitle ="'if($variable=='input'){ title#1 }else { title#2 };'";
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生成的元标题最近在同一个文件中用于创建页面标题(<title></title>)
但它不断产生标题
if($variable=='input'){ title#1 }else { title#2 };
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(整个if语句作为一个整体.它不识别if语句.它将语句视为纯文本.)
我在句子里做错了什么?
我试图找出前3个小时的这个错误.通过搜索,我已经知道这个错误与内存管理有关,但我没有弄清楚我做错了什么.
我已经宣布了这4个标签:
@interface InteractiveHistory : UITableViewController {
UILabel *date;
UILabel *startTime;
UILabel *cal;
UILabel *duration;
}
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然后创建属性并合成它们.在viewDidLoad中,我初始化了所有这样的:
date = [[UILabel alloc] init];
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我也在dealloc()方法中释放它们.
我想要做的是使用这4个标签在表格的单元格中写入一些文本.文本将取决于indexPath.
在cellForRowAtIndexPath方法中,只显示2个标签并指示错误行:
- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
static NSString *CellIdentifier = @"Cell";
UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier];
if (cell == nil) {
cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleSubtitle reuseIdentifier:CellIdentifier] autorelease];
NSLog(@"constructing cell");
//set up labels text
date = [NSString stringWithFormat:@"%@",[[int_data objectAtIndex:indexPath.row] objectForKey:@"Date"]];
NSLog(@"%@",date.text); //App crashes at this line
[date setTag:1];
startTime = [NSString stringWithFormat:@"%@",[[int_data objectAtIndex:indexPath.row] objectForKey:@"Start …Run Code Online (Sandbox Code Playgroud)