【发布时间】:2020-04-03 00:35:23
【问题描述】:
这是It’s OK to Ask and Answer Your Own Questions。 我研究了这个问题,发现结果很奇怪,并发布了我的发现。
从strtoul():1、ULONG_MAX 或什么的“非常负”字符串应该返回什么值?
strtol()
对于表示数值的字符串,如"123",strtol() 的行为与预期一致。字符串(蓝色)[LONG_MIN ... LONG_MAX] 转换为其预期的long 值。 LONG_MAX 上方的字符串(黄色)转换为LONG_MAX 并设置errno==ERANGE。 LONG_MIN 下面的字符串转换为LONG_MIN 并设置errno==ERANGE。
strtoul()正面
对于表示数值的字符串,如"123",strtoul() 的行为也符合预期。字符串(红色)[0 ... ULONG_MAX] 转换为其预期的 unsigned long 值。 ULONG_MAX 上方的字符串(绿色)转换为ULONG_MAX 并设置errno==ERANGE。
如果主题序列以减号开头,则转换产生的值被取反(在返回类型中)。 C17dr § 7.22.1.4 5
strtoul()否定
对于字符串(红色)[-ULONG_MAX ... -1],转换否定正转换并添加 ULONG_MAX + 1(就像将负值分配给 unsigned 的典型分配)并且不设置 errno .这有点令人惊讶,但这就是规范的定义。
strtoul() 非常消极 - 问题
对于小于-ULONG_MAX 的字符串(绿色),我希望将转换处理为上述较小的负值:转换否定正转换(ULONG_MAX 由于溢出)并添加ULONG_MAX + 1。预期结果1(或者可能是0)和errno == ERANGE。然而,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
【问题讨论】:
标签: c