[英]Printing a void* variable in C
Hi all I want to do a debug with printf. 大家好我想用printf进行调试。 But I don't know how to print the "out" variable.
但我不知道如何打印“out”变量。
Before the return, I want to print this value, but its type is void* . 在返回之前,我想打印此值,但其类型为void *。
int
hexstr2raw(char *in, void *out) {
char c;
uint32_t i = 0;
uint8_t *b = (uint8_t*) out;
while ((c = in[i]) != '\0') {
uint8_t v;
if (c >= '0' && c <= '9') {
v = c - '0';
} else if (c >= 'A' && c <= 'F') {
v = 10 + c - 'A';
} else if (c >= 'a' || c <= 'f') {
v = 10 + c - 'a';
} else {
return -1;
}
if (i%2 == 0) {
b[i/2] = (v << 4);
printf("c='%c' \t v='%u' \t b[i/2]='%u' \t i='%u'\n", c,v ,b[i/2], i);}
else {
b[i/2] |= v;
printf("c='%c' \t v='%u' \t b[i/2]='%u' \t i='%u'\n", c,v ,b[i/2], i);}
i++;
}
printf("%s\n", out);
return i;
}
How can I do? 我能怎么做? Thanks.
谢谢。
printf("%p\n", out);
是打印(void*)
指针的正确方法。
This: 这个:
uint8_t *b = (uint8_t*) out;
implies that out
is in fact a pointer to uint8_t
, so perhaps you want to print the data that's actually there. 暗示
out
实际上是指向uint8_t
,所以也许你想要打印实际存在的数据。 Also note that you don't need to cast from void *
in C, so the cast is really pointless. 另请注意,您不需要在C中使用
void *
进行强制转换,因此强制转换非常无意义。
The code seems to be doing hex to binary conversion, storing the results at out
. 代码似乎是进行十六进制到二进制转换,将结果存储
out
。 You can print the i
generated bytes by doing: 您可以通过执行以下操作打印
i
生成的字节:
int j;
for(j = 0; j < i; ++j)
printf("%02x\n", ((uint8_t*) out)[j]);
The pointer value itself is rarely interesting, but you can print it with printf("%p\\n", out);
指针值本身很少有趣,但您可以使用
printf("%p\\n", out);
打印它printf("%p\\n", out);
. 。 The
%p
formatting specifier is for void *
. %p
格式说明符用于void *
。
The format specifier for printing void pointers using printf
in C is %p
. 在C中使用
printf
打印void指针的格式说明符是%p
。 What usually gets printed is a hexadecimal representation of the pointer (although the standard says simply that it is an implementation defined character sequence defining a pointer). 通常打印的是指针的十六进制表示(尽管标准简单地说它是定义指针的实现定义的字符序列)。
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