CGo 将 go 字符串转换为 *C.uchar

Question

var originalMsg *C.uchar
C.ecall_pay_w(8, 10, &originalMsg, &signature)
originalMsgStr := fmt.Sprintf("%c", originalMsg)

//Todo convert originalMstStr to same value with originalMsg

i have to convert go str(originalMsgStr) to *C.uchar type which is same value with originalMsg. How can i do it?

Answer 1

You get a C-string back from your call to C.ecall_pay_w and want to convert that C-string to a Go-string. You can do this by manually following the C-string until you reach the terminating 0 .

Assuming that:

• There is a terminating 0 at the end of the C-string
• The C-string is encoded as ASCII, so every byte represents an ASCII character (in the range [0..127]). This means it is both ASCII and UTF-8 at the same time because UTF-8 is backward compatible to ASCII.

Then your solution could be this:

    func convertCStringToGoString(c *C.uchar) string {
        var buf []byte
        for *c != 0 {
            buf = append(buf, *c)
            c = (*C.uchar)(unsafe.Pointer(uintptr(unsafe.Pointer(c)) + 1))
        }
        return string(buf)
    }

Note that doing "unsafe" things like this in Go is cast-heavy. That was done on purpose by the Go authors. You need to convert to unsafe.Pointer before you can convert to uintptr . The uintptr can be added to ( + 1 ) while the unsafe.Pointer does not support that. These are the reasons for that much casting.

Answer 2

I do not know Go in much detail, but do not forget that in C the *C.uchar would be something like unsigned char * which is often used to reference a string (Null-terminated array of characters).

Here you use fmt.Sprintf("%c", originalMsg) , with %c which expects a single char, so apart from the language detail on how you would cast the resulting string to a *C.uchar , you most probably have lost content already.

%c the character represented by the corresponding Unicode code point

From https://golang.org/pkg/fmt/#hdr-Printing