What would be the canonical (if any, otherwise any reliable) way of detecting if the JIT engine is available in a portable way? For example, Xamarin.iOS does not support JIT since the iOS platform enforces DEP. Mono is quite good at bridging the gap by interpreting most things like lambda expressions these days, but some thing are not (properly) implemented, leading to runtime exceptions that
如果JIT引擎是以便携式方式提供的,那么规范(如果有的话,否则任何可靠的)方法将会是什么? 例如,Xamarin.iOS不支持JIT,因为iOS平台执行DEP。 Mono现在很善于通过解释lambda表达式等大多数东西来弥补差距,但有些东西没有(正确)执行,导致运行时异常严重损害性能。 如果可能的话,我想从共享的可移植类库中进行检测。 你可以尝试执行一个你知道在你的AoT代码上失败但不在JIT上的操作(例如动态创建一个类型),并使
This question already has an answer here: How do you set, clear, and toggle a single bit? 26 answers To set bit 0 to one, use int_var | 1 To reset bit 0 to zero, use int_var & ~1 The operation in your example doesn't seem consistent: 10000011 10000010 & ~1 00000001 xor 00001011 00001011 nop 00000000 xor 01110011 01110010 & ~1 00000001 xor
这个问题在这里已经有了答案: 你如何设置,清除和切换一个位? 26个答案 要将第0位设置为1,请使用 int_var | 1 要将位0重置为零,请使用 int_var & ~1 您的示例中的操作看起来并不一致: 10000011 10000010 & ~1 00000001 xor 00001011 00001011 nop 00000000 xor 01110011 01110010 & ~1 00000001 xor 11101100 11101101 & ~1 00000001 xor 11110101
This question already has an answer here: How do you set, clear, and toggle a single bit? 26 answers If you want to set a bit at third place from the right : Y : 01001000 1 << 2 : 00000100 Y | (1 << 2) : 01001100 The | is OR, bits are set to 1 if any is 1. If you want to remove the bit : 1 << 2 : 00000100 ~(1 << 2) : 11111011 The ~ is
这个问题在这里已经有了答案: 你如何设置,清除和切换一个位? 26个答案 如果你想在右侧排第三位: Y : 01001000 1 << 2 : 00000100 Y | (1 << 2) : 01001100 The | is OR, bits are set to 1 if any is 1. 如果你想删除该位: 1 << 2 : 00000100 ~(1 << 2) : 11111011 The ~ is NOT, bits are inversed Y : 01001100 Y & ~(1 << 2) :
This question already has an answer here: How do you set, clear, and toggle a single bit? 26 answers Certainly. What you want is to perform a logical XOR with 00000001 (this is called a mask): MASK INPUT OUTPUT 00000001 XOR 00000000 = 00000001 00000001 XOR 00000001 = 00000000 This also makes it possible to toggle more than one bit, eg, if your mask were 00001001 : MASK
这个问题在这里已经有了答案: 你如何设置,清除和切换一个位? 26个答案 当然。 你想要的是与00000001执行逻辑异或(这称为掩码): MASK INPUT OUTPUT 00000001 XOR 00000000 = 00000001 00000001 XOR 00000001 = 00000000 这也可以切换多个位,例如,如果你的掩码是00001001 : MASK INPUT OUTPUT 00001001 XOR 00000000 = 00001001 00001001 XOR 00001001 = 00000000 00001001 XOR 000010
This question already has an answer here: How do you set, clear, and toggle a single bit? 26 answers You need to calculate a byte offset and a bitmask within that byte. Set the bit: bitwise OR with mask Clear the bit: bitwise AND with complement of mask Read the bit: return bitwise AND of byte and mask The code: void set_bit(char *buf, int bit, int val) { int byte = bit / 8;
这个问题在这里已经有了答案: 你如何设置,清除和切换一个位? 26个答案 您需要计算该字节内的字节偏移量和位掩码。 设置位:按位或屏蔽 清除该位:用补码掩码按位与 读取该位:返回按位与字节和掩码 代码: void set_bit(char *buf, int bit, int val) { int byte = bit / 8; char mask = 1 << (bit % 8); if (val) buf[byte] |= mask; else buf[byte] &= ~mask; } i
Possible Duplicate: How do you set, clear and toggle a single bit in C? Can some one help me how to toggle a bit at ith position. One way is to ((n>>i)^1) < < i. Are there any other ways ? n ^= 1U << i很容易,不是吗? 你可以做pow(2, i) ^ n
可能重复: 你如何设置,清除并切换C中的一个位? 有人能帮助我如何在第i个位置切换一下。 一种方法是((n >> i)^ 1)<<i。 还有其他方法吗? n ^= 1U << i很容易,不是吗? 你可以做pow(2, i) ^ n
Possible Duplicate: How do you set, clear and toggle a single bit in C? I want to create an assembler , so I need to assign 32 bits bit by bit or field by field to create 32 bit opcode from assembly ... how can I do this in C ? how can I assign bits in integer ? can this be done ? You can declare these two macros to help you: #define Set_Bit(IntValue, BitNumber) IntValue = IntValue | (1&
可能重复: 你如何设置,清除并切换C中的一个位? 我想创建一个汇编程序,所以我需要逐位或逐字段地分配32位,以从程序集中创建32位操作码...我如何在C中执行此操作? 我如何分配整数位? 这可以做到吗? 你可以声明这两个宏来帮助你: #define Set_Bit(IntValue, BitNumber) IntValue = IntValue | (1<<BitNumber) #define Clr_Bit(IntValue, BitNumber) IntValue = IntValue & (~((1) << (BitNumber))
This question already has an answer here: How do you set, clear, and toggle a single bit? 26 answers First you have to make sure to use an integer type that has the same size as a pointer. On most 64-bit platforms, int s are 32-bit and pointers are 64-bit, so you'll corrupt the pointer when casting to an int . size_t usually does the job, except for some exotic memory models. Then I&
这个问题在这里已经有了答案: 你如何设置,清除和切换一个位? 26个答案 首先,您必须确保使用与指针大小相同的整数类型。 在大多数64位平台上, int是32位的,指针是64位的,所以在转换为int时会损坏指针。 除了一些奇特的内存模型外, size_t通常可以完成这项工作。 然后,我建议使用一个联合,它允许修改一个指针的位而不需要任何强制转换: union { size_t *pointer; size_t integer; } u; u.pointer =
This question already has an answer here: How do you set, clear, and toggle a single bit? 26 answers This should help http://en.wikipedia.org/wiki/Bitwise_operations_in_C If you need working example let me know but i encourage you to figure it out on your own. unsigned char my_byte = 0x3B; // 0b00111011 // clear the bits my_byte &= 0xE3; // set the bits my_byte |= 0x14; You'll
这个问题在这里已经有了答案: 你如何设置,清除和切换一个位? 26个答案 这应该有助于http://en.wikipedia.org/wiki/Bitwise_operations_in_C 如果你需要工作的例子,让我知道,但我鼓励你自己弄清楚。 unsigned char my_byte = 0x3B; // 0b00111011 // clear the bits my_byte &= 0xE3; // set the bits my_byte |= 0x14; 你会发现许多人对如何编写0xE3和0x14有许多不同的偏好。 有些人喜欢移位,但最终这是应
This question already has an answer here: How do you set, clear, and toggle a single bit? 26 answers You just have to replace the logical OR with a logical AND operation. You would use the & operator for that: pt = pt & ~(1 << i); You have to invert your mask because logical AND ing with a 1 will maintain the bit while 0 will clear it... so you'd need to specify a 0 in t
这个问题在这里已经有了答案: 你如何设置,清除和切换一个位? 26个答案 您只需用逻辑AND操作替换逻辑OR 。 你可以使用&运算符: pt = pt & ~(1 << i); 你必须反转你的掩码,因为逻辑AND与1将保持该位,而0将清除它...所以你需要在你想清除的位置指定一个0 。 具体来说,做1 << i会给你一个000...010..000的掩码,其中1是在你想要的位的位置, 000...010..000给出111...101...111 。 逻辑AND与这