是否有一种简单的方法可以转义printf()
函数中的所有特殊字符?
我想知道如何做到这一点的原因是因为我打印了许多字符,其中可能包含特殊字符,例如空字符(\0
)和哔声字符,我只是想看到字符串的内容。
目前我使用以下代码
适用于空字符。逃脱所有特殊角色的最简单方法是什么?
int length;
char* data = GetData( length ); // Fills the length as reference
for( int i = 0; i < length; i++ )
{
char c = data[ i ];
printf( "%c", ( c == 0 ? '\\0' : data[ i ] ) );
}
答案 0 :(得分:3)
首先,'\\0'
是一个双字符的字面值,它应该是一个双字符字符串。至于将所有特殊字符打印为转义码,您需要更多代码:
switch (data[i])
{
case '\0':
printf("\\0");
break;
case '\n':
printf("\\n");
break;
/* Etc. */
default:
/* Now comes the "hard" part, because not all characters here
* are actually printable
*/
if (isprint(data[i]))
printf("%c", data[i]); /* Printable character, print it as usual */
else
printf("\\x%02x", data[i]); /* Non-printable character, print as hex value */
break;
}
答案 1 :(得分:2)
使用isprint
库函数确定字符是否可打印:
#include <ctype.h>
...
if (isprint(data[i]))
printf(" %c", data[i]); // prints character
else
printf(" %d", data[i]); // prints code value for character
答案 2 :(得分:0)
如果代码需要用 no 歧义编写,请使用C语法:
#include <ctype.h>
#include <string.h>
#include <stdio.h>
void EscapePrint(int ch) {
// Delete or adjust these 2 arrays per code's goals
// All simple-escape-sequence C11 6.4.4.4
static const char *escapev = "\a\b\t\n\v\f\r\"\'\?\\";
static const char *escapec = "abtnvfr\"\'\?\\";
char *p = strchr(escapev, ch);
if (p && *p) {
printf("\\%c", escapec[p - escapev]);
} else if (isprint(ch)) {
fputc(ch, stdout);
} else {
// Use octal as hex is problematic reading back
printf("\\%03o", ch);
}
}
void EscapePrints(const char *data, int length) {
while (length-- > 0) {
EscapePrint((unsigned char) *data++);
}
}
或者,代码可以
void EscapePrint(char sch) {
int ch = (unsigned char) sch;
...
}
void EscapePrints(const char *data, int length) {
while (length-- > 0) {
EscapePrint(*data++);
}
}
要使用十六进制转义序列或缩短八进制转义序列,代码需要确保 next 字符不会产生歧义。上述代码中不会出现这种复杂情况,因为它使用3位八进制转义序列。修改后的代码类似于:
} else {
if ((ch == 0) && (nextch < '0' || nextch > '7')) {
fputs("\\0", stdout);
}
else if (!isxdigit((unsigned char) nextch)) {
printf("\\x%X", ch);
}
else {
// Use octal as hex is problematic reading back
printf("\\%03o", ch);
}
}
答案 3 :(得分:0)
#include <stdio.h>
#include <ctype.h>
/* Converts a buffer of specified lenth to
* ASCII representation as it was a C string literal.
* Returns how much bytes from source was processed
* (ideally ret == src_sz)
*/
int binbuf_to_escaped_C_literal(const char *src_buf, size_t src_sz, char *dst_str, size_t dst_sz)
{
const char *src = src_buf;
char *dst = dst_str;
while (src < src_buf + src_sz)
{
if (*src == '\\')
{
*dst++ = '\\';
*dst++ = *src++;
}
else if (isprint(*src))
{
*dst++ = *src++;
}
else
{
switch(*src)
{
case '\n':
*dst++ = '\\';
*dst++ = 'n';
break;
case '\r':
*dst++ = '\\';
*dst++ = 'r';
break;
case '\t':
*dst++ = '\\';
*dst++ = 't';
break;
case '\0':
*dst++ = '\\';
*dst++ = '0';
break;
default:
sprintf(dst, "0x%x", *src);
dst += 4;
}
src++;
}
// next iteration requires up to 5 chars in dst buffer, for ex. "0xab\0"
if (dst > (dst_str + dst_sz - 5)) {
break;
}
}
*dst = '\0';
return src - src_buf;
}
int main(int argc, char **argv)
{
const char binbuf[] = "strange \n\r\t\0\0\0\0\0\\\\ string";
size_t sz = sizeof(binbuf) - 1; // drop trailing nul terminator
char escaped[128];
if (binbuf_to_escaped_C_literal(binbuf, sz, escaped, sizeof(escaped)) != sz) {
fprintf(stderr, "Destination string buffer is too small\n");
return 1;
}
printf("Escaped: %s\n", escaped);
// $ ./escape-binary-buf //
// Escaped: strange \n\r\t\0\0\0\0\0\\\\ string //
return 0;
}