verilog fwrite 输出字节 [英] verilog fwrite output bytes

问题描述

我知道如果我在verilog中输出一个二进制文件，那么我可以使用下面的verilog IO标准函数:

I know that if I am outputting a binary file in verilog, then I can use the following verilog IO standard function:

$fwrite(fd,"%u",32'hABCDE124); 

但是上面的命令将 4 字节的数据写入文件.如果我要写入的二进制数据只有一字节、两字节或三字节怎么办?我该怎么做?

But the above command writes 4-byte data into the file. What if the binary data that I want to write is only one-byte, two-bytes or three-bytes? How can I do this?

例如，我知道以下不会做我想要的:

For example, I know the following won't do what I want:

$fwrite(fd,"%u",8'h24);
$fwrite(fd,"%u",16'hE124);
$fwrite(fd,"%u",24'hCDE124);

有什么办法可以将非 4 字节的多个数据写入文件?

Is there any way that I can write a non 4-byte multiple data into a file?

谢谢，

--鲁迪

推荐答案

我建议使用另一个版本的 dave_59 的答案.关键是使用多个%c.

I suggest another version of dave_59's answer. The key is using multiple %c.

wire [7:0] byte;
wire [15:0] two_bytes;
wire [23:0] three_bytes;
----
assign byte = 8'h24;
assign two_bytes = 16'hE124;
assign three_bytes = 24'hCDE124;
----
$fwrite(fd_s,"%c",byte);
$fwrite(fd_s,"%c%c",two_bytes[15-:8],two_bytes[7-:8]);
$fwrite(fd_s,"%c%c%c",three_bytes[23-:8],three_bytes[15-:8],three_bytes[7-:8]);

使用 %u 时 - 请注意您的系统字节顺序，它默认使用本机.

When using %u - be aware of Your system byte ordering, it uses native by default.

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