这是一提出并回答自己的问题是可以的。
我研究了这个问题,发现结果很奇怪,并发布了我的发现。
“非常负”的字符串应该返回什么值strtoul()
: 1
, ULONG_MAX
或者是什么?
strtol()
对于表示数值的字符串,例如"123"
, strtol()
行为符合预期。琴弦(蓝色)[LONG_MIN ... LONG_MAX]
转换成他们的预期long
价值。上面的绳子(黄色)LONG_MAX
转换成LONG_MAX
并设置errno==ERANGE
。下面的字符串LONG_MIN
转换成LONG_MIN
并设置errno==ERANGE
.
strtoul()
积极的
对于表示数值的字符串,例如"123"
, strtoul()
也按预期运行。弦(红)[0 ... ULONG_MAX]
转换成他们的预期unsigned long
价值。上面的字符串(绿色)ULONG_MAX
转换成ULONG_MAX
并设置errno==ERANGE
.
如果主题序列以减号开头,则转换产生的值将被取反(在返回类型中)。 C17dr § 7.22.1.4 5
strtoul()
消极的
对于字符串(红色)[-ULONG_MAX ... -1]
,转换否定正转换并添加ULONG_MAX + 1
(就像典型的将负值分配给unsigned)并且不设置errno
。这有点令人惊讶,但这就是规范的定义方式。
strtoul()
非常消极 - 问题
对于字符串(绿色)小于-ULONG_MAX
,我希望转换像上述较小的负值一样处理:转换否定正转换(ULONG_MAX
由于溢出)并添加ULONG_MAX + 1
。预期结果1
(或者可能0
) with errno == ERANGE
。还在锻炼strtoul()
导致ULONG_MAX
.
什么是正确的?
测试代码
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
long strtol_test(const char *s, int base) {
printf("\n");
int width = snprintf(NULL, 0, "%ld", LONG_MIN);
printf("base:%2d \"%s\"\n", base, s);
char *endptr_signed;
errno = 0;
long val_signed = strtol(s, &endptr_signed, base);
int errno_signed = errno;
char *endptr_unsigned;
errno = 0;
unsigned long val_unsigned = strtoul(s, &endptr_unsigned, base);
int errno_unsigned = errno;
if (val_signed < 0 || (unsigned long) val_signed != val_unsigned
|| endptr_signed != endptr_unsigned || errno_signed != errno_unsigned) {
printf(" signed val:%*ld end:%2td e:%s\n", width, val_signed,
endptr_signed - s, strerror(errno_signed));
printf("unsigned val:%*lu end:%2td e:%s\n", width, val_unsigned,
endptr_unsigned - s, strerror(errno_unsigned));
return 1;
}
printf(" both val:%*ld end:%2td e:%s\n", width, val_signed,
endptr_signed - s, strerror(errno_signed));
return 0;
}
int main() {
char s[][50] = {"-ULONG_MAX1", "-ULONG_MAX", "LONG_MIN1", "LONG_MIN", "-1",
"-0", "42", "LONG_MAX", "LONG_MAX1", "ULONG_MAX", "ULONG_MAX1", "x"};
snprintf(s[0], sizeof *s, "-%lu", ULONG_MAX);
s[0][strlen(s[0]) - 1]++;
snprintf(s[1], sizeof *s, "-%lu", ULONG_MAX);
snprintf(s[2], sizeof *s, "%ld", LONG_MIN);
s[2][strlen(s[2]) - 1]++;
snprintf(s[3], sizeof *s, "%ld", LONG_MIN);
snprintf(s[7], sizeof *s, "%ld", LONG_MAX);
snprintf(s[8], sizeof *s, "%ld", LONG_MAX);
s[8][strlen(s[8]) - 1]++;
snprintf(s[9], sizeof *s, "%lu", ULONG_MAX);
snprintf(s[10], sizeof *s, "%lu", ULONG_MAX);
s[10][strlen(s[10]) - 1]++;
strcpy(s[11], s[0]);
s[11][strlen(s[11]) - 1]++;
int n = sizeof s / sizeof s[0];
for (int i = 0; i < n; i++) {
strtol_test(s[i], 0);
}
}
样本输出
base: 0 "-18446744073709551616"
signed val:-9223372036854775808 end:21 e:Numerical result out of range
unsigned val:18446744073709551615 end:21 e:Numerical result out of range
base: 0 "-18446744073709551615"
signed val:-9223372036854775808 end:21 e:Numerical result out of range
unsigned val: 1 end:21 e:No error
base: 0 "-9223372036854775809"
signed val:-9223372036854775808 end:20 e:Numerical result out of range
unsigned val: 9223372036854775807 end:20 e:No error
base: 0 "-9223372036854775808"
signed val:-9223372036854775808 end:20 e:No error
unsigned val: 9223372036854775808 end:20 e:No error
base: 0 "-1"
signed val: -1 end: 2 e:No error
unsigned val:18446744073709551615 end: 2 e:No error
base: 0 "-0"
both val: 0 end: 2 e:No error
base: 0 "42"
both val: 42 end: 2 e:No error
base: 0 "9223372036854775807"
both val: 9223372036854775807 end:19 e:No error
base: 0 "9223372036854775808"
signed val: 9223372036854775807 end:19 e:Numerical result out of range
unsigned val: 9223372036854775808 end:19 e:No error
base: 0 "18446744073709551615"
signed val: 9223372036854775807 end:20 e:Numerical result out of range
unsigned val:18446744073709551615 end:20 e:No error
base: 0 "18446744073709551616"
signed val: 9223372036854775807 end:20 e:Numerical result out of range
unsigned val:18446744073709551615 end:20 e:Numerical result out of range
base: 0 "-18446744073709551617"
signed val:-9223372036854775808 end:21 e:Numerical result out of range
unsigned val:18446744073709551615 end:21 e:Numerical result out of range