将float数组写为反向字节

时间:2014-10-31 10:15:30

标签: ios byte nsdata uint8t

我将使用iOS objective-C创建所需输出的字节数组。该方法是从静态浮点数组转换为int8_t数组字节数组。说到实现,我发现float数组中每个float的所有字节都按顺序颠倒过来。输出显示为实际输出。您能否告诉我如何转换每个字节并显示为所需输出?以下是我的工作:

 float floatArray[5] = {100.0 , 10.0 , 10.0 , 10.0 , 10.0 };
    NSUInteger lengthN = sizeof(floatArray) ;
    NSLog(@" length %lu" , (unsigned long)lengthN);


    int8_t oneByte = lengthN;
    int8_t preffix[4]  = {0x26, 0x24, 0x61 , oneByte };
     //  NSArray *charArray = arry; //20d = 14h
    //char arry[4]={ 0x26, 0x24, 0x61  , oneByte };
    int8_t data[lengthN + 5];


    memcpy (data, (int8_t *) &preffix, sizeof(preffix));

    memcpy (data+4, (int8_t *) &floatArray,  lengthN );

    int length = (int)lengthN + 5;
    int checkSum = 119 + 97 + (int)lengthN ;

    for(int  i = 4 ; i  < lengthN * 4 ; i *=4 ){
        [self swap: data[ 4*i +3] : data [4*i+ 0]];
        [self swap: data[ 4*i +2] : data [4*i+ 1]];
        [self swap: data[ 4*i +1] : data [4*i+ 2]];
        [self swap: data[ 4*i +0] : data [4*i+ 3]];

    }



- (void)swap:(int8_t)a :(int8_t)b {
    a ^= b;
    b ^= a;
    a ^= b;

}

实际输出

2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal -56
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 66
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.666 marker[2770:1000346] sdjhasdhal 32
2014-10-31 18:09:21.666 marker[2770:1000346] sdjhasdhal 65
2014-10-31 18:09:21.666 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 32
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 65
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 32
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 65
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.668 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.668 marker[2770:1000346] sdjhasdhal 32
2014-10-31 18:09:21.668 marker[2770:1000346] sdjhasdhal 65

期望的输出

2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 66
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal -56
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 65
2014-10-31 18:09:21.665 marker[2770:1000346] sdjhasdhal 32
2014-10-31 18:09:21.666 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.666 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.666 marker[2770:1000346] sdjhasdhal 65
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 32
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 65
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 32
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.667 marker[2770:1000346] sdjhasdhal 65
2014-10-31 18:09:21.668 marker[2770:1000346] sdjhasdhal 32
2014-10-31 18:09:21.668 marker[2770:1000346] sdjhasdhal 0
2014-10-31 18:09:21.668 marker[2770:1000346] sdjhasdhal 0

1 个答案:

答案 0 :(得分:0)

你正在使用一个可怕的,可怕的,可怕的技巧来交换字节。在代码审查中,您没有任何机会通过代码审查。

更糟糕的是你交换字节0和3 两次,并且相同的字节1和2.猜猜当你交换两次时会发生什么?没有。

您也在假设处理器的字节顺序(因为实际上,您不希望反转字节,您希望将它们置于正确的顺序中)。为此,memcpy从float到uint32_t,然后通过按照你想要的顺序将uint32_t的内容移动24,16,8和0位来提取四个字节。然后处理器的字节顺序无关紧要。

PS。我对你的字节交换代码感到很生气,我甚至没有意识到你没有通过指针......

PPS。大多数人讨厌变量名中的拼写错误,比如“preffix”。

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