uint32-t

How can i convert a uint32_t to a char* type

由 偶尔善良 提交于 2020-12-05 07:25:46
问题 Hello i am using an Arduino UNO with an adafruit shield to display score values but the function used to display scores only accepts char* values and the score itself can occupy up to 6 digits(000,000 to 999,999). i have tried using sprint() but i have had no luck since the screen will flicker like crazy. i believe the problem to be that chars only hold a certain number of bytes that could not fit a 32 bit int but i would think their is a way around this. draw text is the function used by the

How can i convert a uint32_t to a char* type

由 扶醉桌前 提交于 2020-12-05 07:24:38
问题 Hello i am using an Arduino UNO with an adafruit shield to display score values but the function used to display scores only accepts char* values and the score itself can occupy up to 6 digits(000,000 to 999,999). i have tried using sprint() but i have had no luck since the screen will flicker like crazy. i believe the problem to be that chars only hold a certain number of bytes that could not fit a 32 bit int but i would think their is a way around this. draw text is the function used by the

How can i convert a uint32_t to a char* type

由 拥有回忆 提交于 2020-12-05 07:24:32
问题 Hello i am using an Arduino UNO with an adafruit shield to display score values but the function used to display scores only accepts char* values and the score itself can occupy up to 6 digits(000,000 to 999,999). i have tried using sprint() but i have had no luck since the screen will flicker like crazy. i believe the problem to be that chars only hold a certain number of bytes that could not fit a 32 bit int but i would think their is a way around this. draw text is the function used by the

What is *(uint32_t*)

由 自古美人都是妖i 提交于 2020-06-07 12:07:45
问题 I have a hard time understanding *(uint32_t*) . Let's say that I have: uint32_t* ptr; uint32_t num *(uint32_t*)(ptr + num); //what does this do? does it 回答1: uint32_t is a numeric type that guarantees 32 bits, the value is unsigned, meaning that the range of values goes from 0 to 2 32 - 1. This uint32_t* ptr; declares a pointer of type uint32_t* , but the pointer is uninitialized, that is, the pointer does not point to anywhere in particular. Trying to access memory through that pointer will

What is *(uint32_t*)

由 徘徊边缘 提交于 2020-06-07 12:06:29
问题 I have a hard time understanding *(uint32_t*) . Let's say that I have: uint32_t* ptr; uint32_t num *(uint32_t*)(ptr + num); //what does this do? does it 回答1: uint32_t is a numeric type that guarantees 32 bits, the value is unsigned, meaning that the range of values goes from 0 to 2 32 - 1. This uint32_t* ptr; declares a pointer of type uint32_t* , but the pointer is uninitialized, that is, the pointer does not point to anywhere in particular. Trying to access memory through that pointer will

'uint32_t' does not name a type

由 大兔子大兔子 提交于 2019-12-31 08:15:27
问题 I'm trying to compile a C++ software package that was written in 2007 and I'm getting this error: error: ‘uint32_t’ does not name a type This is happening in 64-bit Ubuntu using g++ 4.5.2. It compiles fine on 64-bit CentOS using g++ 4.1.2. Is there an #include or a compiler flag that I'm missing? Or, should I use typedef to assign uint32_t to a size_t or maybe an unsigned int ? 回答1: You need to include stdint.h #include <stdint.h> 回答2: You need to #include <cstdint> , but that may not always

Difference between uint32 and uint32_t [duplicate]

由 拥有回忆 提交于 2019-12-20 08:23:18
问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Difference between different integer types What is the difference between uint32 and uint32_t in C/C++? Are they OS dependent? In which case should I use one or another? Thanks 回答1: uint32_t is standard, uint32 is not. That is, if you include <inttypes.h> or <stdint.h> , you will get a definition of uint32_t . uint32 is a typedef in some local code base, but you should not expect it to exist unless you define it

When should I use UINT32_C(), INT32_C(),… macros in C?

由 人盡茶涼 提交于 2019-12-06 18:33:25
问题 I switched to fixed-length integer types in my projects mainly because they help me think about integer sizes more clearly when using them. Including them via #include <inttypes.h> also includes a bunch of other macros like the printing macros PRIu32 , PRIu64 ,... To assign a constant value to a fixed length variable I can use macros like UINT32_C() and INT32_C() . I started using them whenever I assigned a constant value. This leads to code similar to this: uint64_t i; for (i = UINT64_C(0);

When should I use UINT32_C(), INT32_C(),… macros in C?

由 自作多情 提交于 2019-12-04 22:30:30
I switched to fixed-length integer types in my projects mainly because they help me think about integer sizes more clearly when using them. Including them via #include <inttypes.h> also includes a bunch of other macros like the printing macros PRIu32 , PRIu64 ,... To assign a constant value to a fixed length variable I can use macros like UINT32_C() and INT32_C() . I started using them whenever I assigned a constant value. This leads to code similar to this: uint64_t i; for (i = UINT64_C(0); i < UINT64_C(10); i++) { ... } Now I saw several examples which did not care about that. One is the

'uint32_t' does not name a type

由 隐身守侯 提交于 2019-12-02 17:14:10
I'm trying to compile a C++ software package that was written in 2007 and I'm getting this error: error: ‘uint32_t’ does not name a type This is happening in 64-bit Ubuntu using g++ 4.5.2. It compiles fine on 64-bit CentOS using g++ 4.1.2. Is there an #include or a compiler flag that I'm missing? Or, should I use typedef to assign uint32_t to a size_t or maybe an unsigned int ? You need to include stdint.h #include <stdint.h> plasma You need to #include <cstdint> , but that may not always work. The problem is that some compiler often automatically export names defined in various headers or
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