promotions

问题 I was studying about default argument promotions and got stuck at one point. In C 2011 (ISO/IEC 9899:2011), the relevant part seem to be: §6.5.2.2 Function calls ¶6 If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double. These are called the default argument promotions. If the number of arguments does not equal the number of parameters, the

问题 I was studying about default argument promotions and got stuck at one point. In C 2011 (ISO/IEC 9899:2011), the relevant part seem to be: §6.5.2.2 Function calls ¶6 If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double. These are called the default argument promotions. If the number of arguments does not equal the number of parameters, the

问题 Apple doesn't offer promotional codes for in-app purchases. What's the best way to let users try the features or content unlocked by in-app purchases for free, while complying with Apple's Developer Guidelines? The idea is to allow a special set of users (reviewers, key fans, etc.) to access the content or features offered as in-app purchases without paying. Examples of apps that worked around this limitation would be much appreciated. 回答1: You could submit a version of your application that

问题 Please let me know what is the best way to determine composition of young generation memory promoted to old generation, after each young GC event? Ideally I would like to know class names which are responsible say, for 80% of heap in each "young gen -> old gen" promotion chunk; Example: I have 600M young gen, each tenure promotes 6M; I want to know which objects compose this 6M. Thank you. 回答1: There is no easy way to do this, however, I have recently been analyzing memory performance of

问题 I have two jobs in Jenkins. First of the name “Build” and the second of the name “Deploy to test environment”. In the first job, tester sets promotion manually, and then only promoted builds can be deployed. In second job I added “Promoted Build Parameter” which generates combobox with promoted builds but I can’t connect the value of this parameter with “Copy artifact from another project” build step. So how can I copy artifacts from the selected promoted build? 回答1: In your deploy project:

问题 I installed the Promoted Build Plugin from Jenkins and now I'm facing some troubles to promote a build from an existing job. Here is the scenario: There is an existing Nightly Build job that runs every night running all the tests and metrics needed; There is an existing Deploy Build that accepts a parameter ${BUILD_NUMBER} and deploys the build that has the corresponding ${BUILD_NUMBER} from the Nightly Build Say the [Nightly Build] ran and successfully built the artifact #39 Now I can just

问题 In the three bitwise left shift code fragments below, it's interesting that examples #2 and #3 are treated differently by Java. In the last example (#3), why does Java decide not to upgrade the compound assignment statement to an int? Does the answer have something to do with Java doing things "inline". Thanks a lot for any comments. byte b = -128; // Eg #1. Expression is promoted to an int, and its expected value for an int is -256. System.out.println(b << 1); b = -128; // Eg #2. Must use a

问题 I'm getting MISRA type errors when I use "%f" specifier for snprintf with a parameter of type float . According to my research, MISRA is correct because "%f" expectes a type of double . Is there a floating point specifier or modifier that will use a float type parameter and not a double ? I'm working on an embedded system and don't want to convert from 32-bit float to 64-bit double just to please the snprintf function. The code prints to the debug / console port and this is the only place

问题 #include "stdio.h" int main() { int x = -13701; unsigned int y = 3; signed short z = x / y; printf("z = %d\n", z); return 0; } I would expect the answer to be -4567. I am getting "z = 17278". Why does a promotion of these numbers result in 17278? I executed this in Code Pad. 回答1: The hidden type conversions are: signed short z = (signed short) (((unsigned int) x) / y); When you mix signed and unsigned types the unsigned ones win. x is converted to unsigned int , divided by 3, and then that

问题 From this other QUESTION they talk about how Bjarne Stroustrup said that just as integral data-types narrower than an int (e.g. short ) are promoted to an int , float s are promoted to a double . However, unlike widening of integrals narrower than an int , floating point promotion does not happen in the same way, but instead, occurs elsewhere. I know that if you were to compute float + double the float would be converted to a double before the binary operator( + ) is applied. However, this is