[英]C converting 20 digit string to number for IBAN validation
I am doing an IBAN validation in C.我正在 C 中进行 IBAN 验证。 For this I have a char* which is something like '2012129431327715102998'.
为此,我有一个 char*,类似于“2012129431327715102998”。 Now I want to check the IBAN by taken the value modulo 97. So I want to do
2012129431327715102998 % 97
.现在我想通过取模 97 的值来检查 IBAN。所以我想做
2012129431327715102998 % 97
。 I have already tried to convert the char* with strtoull but this gives me an out-of-range error.我已经尝试使用 strtoull 转换 char* ,但这给了我一个超出范围的错误。 So my question is: How can I convert this char* to a number where I can do a modulo calculation?
所以我的问题是:如何将此 char* 转换为可以进行模计算的数字? Thanks in advance
提前致谢
A simple way without using additional library is to remember that mathematically : mod(a*b, c) == mod(b * mod(a, c), c)
.一个不使用额外库的简单方法是在数学上记住:
mod(a*b, c) == mod(b * mod(a, c), c)
。 So you can process the number in chunks :因此,您可以分块处理数字:
// suitable for a 32 bits system, can use 8 for a 64 bits one
#define NB 4
/*********************
* Initial is a string containin only digits representing an arbitrary large number
* div in a number < 10000 (because NB is 4)
* ******************/
int large_mod(char *initial, int div) {
char old[1 + (NB * 2)] = ""; // enough room for a remainder and next chunk
long val;
for (unsigned i=0; i<strlen(initial); i+= NB) {
strncat(old, initial + i, NB); // add the new chunk
val = atol(old) % div; // compute the remainder
sprintf(old, "%ld", val); // keep it for next chunk
// printf("%ld ", val); // uncomment for debugging
}
return (int) val;
}
For 2012129431327715102998 % 97, it gives as expected 53.对于 2012129431327715102998 % 97,它给出了预期的 53。
You can write a custom function for this.您可以为此编写自定义 function。 Applying the modulo operator on partial sums, you can convert a number of arbitrary length:
对部分和应用模运算符,您可以转换多个任意长度:
#include <stdio.h>
int mod97(const char *s) {
int res = 0;
while (*s >= '0' && *s <= '9') {
res = (res * 10 + (*s++ - '0')) % 97;
}
return res;
}
int main(int argc, char *argv[]) {
for (int i = 1; i < argc; i++) {
printf("%s -> %d\n", argv[i], mod97(argv[i]));
}
return 0;
}
Output: Output:
./mod97 2012129431327715102998
2012129431327715102998 -> 53
This method is simpler and more generic than the one described in the wiki article: computing the modulo 97 of a large number can be achieved by splitting the number in chunks of 9 digits and combining the modulo of these chunks.这种方法比wiki文章中描述的方法更简单、更通用:计算一个大数的模 97 可以通过将数字分成 9 位的块并组合这些块的模来实现。 This splitting is specific to
97
and works because 1000000000 % 97 == 1
.这种拆分特定于
97
并且有效,因为1000000000 % 97 == 1
。 The above method works for any modulo value up to INT_MAX / 10
.上述方法适用于最大为
INT_MAX / 10
的任何模值。
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