long-integer

The maximum string content length quota (8192) has been exceeded while reading XML data

由 这一生的挚爱 提交于 2019-11-26 07:46:24
问题 I\'m trying to pass a large string (24,000 to 50,000 characters) to a self-hosted TCP WCF service. I\'ve upped the maxStringContentLength (everywhere) to 22008192. I read somewhere that I need to change the bindingConfiguration to \"LargeBuffer\" or \"LongFields\" but when I do this: <endpoint address=\"\" binding=\"netTcpBinding\" bindingConfiguration=\"LongFields\" contract=\"ExStreamWCF.IService1\"> or this: <endpoint address=\"\" binding=\"netTcpBinding\" bindingConfiguration=\

What is the difference between “long”, “long long”, “long int”, and “long long int” in C++?

由 三世轮回 提交于 2019-11-26 06:53:23
问题 I am transitioning from Java to C++ and have some questions about the long data type. In Java, to hold an integer greater than 2 32 , you would simply write long x; . However, in C++, it seems that long is both a data type and a modifier. There seems to be several ways to use long : long x; long long x; long int x; long long int x; Also, it seems there are things such as: long double x; and so on. What is the difference between all of these various data types, and do they all have the same

Is `long` guaranteed to be at least 32 bits?

由 和自甴很熟 提交于 2019-11-26 05:56:15
问题 By my reading of the C++ Standard, I have always understood that the sizes of the integral fundamental types in C++ were as follows: sizeof(char) <= sizeof(short int) <= sizeof(int) <= sizeof(long int) I deduced this from 3.9.1/2: There are four signed integer types: “signed char”, “short int”, “int”, and “long int.” In this list, each type provides at least as much storage as those preceding it in the list. Plain ints have the natural size suggested by the architecture of the execution

Java&#39;s L number (long) specification

由 ￡可爱￡侵袭症＋ 提交于 2019-11-26 05:55:46
问题 It appears that when you type in a number in Java, the compiler automatically reads it as an integer, which is why when you type in (long) 6000000000 (not in integer\'s range) it will complain that 6000000000 is not an integer. To correct this, I had to specify 6000000000L . I just learned about this specification. Are there other number specifications like for short, byte, float, double? It seems like these would be good to have because (I assume) if you could specify the number you\'re

How to convert a hexadecimal string to long in java?

由 痞子三分冷 提交于 2019-11-26 05:35:53
问题 I want to convert a hex string to long in java. I have tried with general conversion. String s = \"4d0d08ada45f9dde1e99cad9\"; long l = Long.valueOf(s).longValue(); System.out.println(l); String ls = Long.toString(l); But I am getting this error message: java.lang.NumberFormatException: For input string: \"4d0d08ada45f9dde1e99cad9\" Is there any way to convert String to long in java? Or am i trying which is not really possible!! Thanks! 回答1: Long.decode(str) accepts a variety of formats:

long long in C/C++

由 只谈情不闲聊 提交于 2019-11-26 03:52:21
问题 I am trying this code on GNU\'s C++ compiler and am unable to understand its behaviour: #include <stdio.h>; int main() { int num1 = 1000000000; long num2 = 1000000000; long long num3; //num3 = 100000000000; long long num4 = ~0; printf(\"%u %u %u\", sizeof(num1), sizeof(num2), sizeof(num3)); printf(\"%d %ld %lld %llu\", num1, num2, num3, num4); return 0; } When I uncomment the commented line, the code doesn\'t compile and is giving an error: error: integer constant is too large for long type

wordwrap a very long string

由 生来就可爱ヽ(ⅴ＜●) 提交于 2019-11-26 03:36:02
问题 How can you display a long string, website address, word or set of symbols with automatic line breaks to keep a div width? I guess a wordwrap of sorts. Usually adding a space works but is there a CSS solution such as word-wrap? For example it (very nastily) overlaps divs, forces horizontal scrolling etc. wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww What can I add to the above string to fit

How do I convert Long to byte[] and back in java

由 。_饼干妹妹 提交于 2019-11-26 03:34:23
How do I convert a long to a byte[] and back in Java? I'm trying convert a long to a byte[] so that I will be able to send the byte[] over a tcp connection. On the other side I want to take that byte[] and convert it back into a double. Any tips would be appreciated. Brad Mace public byte[] longToBytes(long x) { ByteBuffer buffer = ByteBuffer.allocate(Long.BYTES); buffer.putLong(x); return buffer.array(); } public long bytesToLong(byte[] bytes) { ByteBuffer buffer = ByteBuffer.allocate(Long.BYTES); buffer.put(bytes); buffer.flip();//need flip return buffer.getLong(); } Or wrapped in a class to

Why do these two multiplication operations give different results?

由 ＊爱你＆永不变心＊ 提交于 2019-11-26 03:19:38
问题 Why do I need to add an \"L\" letter to get the correct long value? And what is the other value? long oneYearWithL = 1000*60*60*24*365L; long oneYearWithoutL = 1000*60*60*24*365; System.out.println(oneYearWithL);//gives correct calculation result : 31536000000 System.out.println(oneYearWithoutL)//gives incorrect calculation result: 1471228928 回答1: long oneYearWithL = 1000*60*60*24*365L; long oneYearWithoutL = 1000*60*60*24*365; Your first value is actually a long (Since 365L is a long , and

How can I check if multiplying two numbers in Java will cause an overflow?

由 帅比萌擦擦* 提交于 2019-11-26 03:08:04
问题 I want to handle the special case where multiplying two numbers together causes an overflow. The code looks something like this: int a = 20; long b = 30; // if a or b are big enough, this result will silently overflow long c = a * b; That\'s a simplified version. In the real program a and b are sourced elsewhere at runtime. What I want to achieve is something like this: long c; if (a * b will overflow) { c = Long.MAX_VALUE; } else { c = a * b; } How do you suggest I best code this? Update: a
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