我正在致力于将DNA序列转化为蛋白质序列。
我已经完成了所有程序,我发现有一个结构的错误。
dna_codon是一种结构,我在它上进行迭代。在第一次迭代中,它显示了正确的结构值,但是从下一个迭代开始,它不会显示在结构中存储的正确值。
这是一个很小的错误,所以不要以为我做任何事情并低估了。我之所以被困在这里,是因为我是C的新来的结构。
代码:
#include <stdio.h>
#include<string.h>
void main()
{
int i, len;
char short_codons[20];
char short_slc[1000];
char sequence[1000];
struct codons
{
char amino_acid[20], slc[20], dna_codon[40];
};
struct codons c1 [20]= {
{"Isoleucine", "I", "ATT, ATC, ATA"},
{"Leucine", "L", "CTT, CTC, CTA, CTG, TTA, TTG"},
{"Valine", "V", "GTT, GTC, GTA, GTG"},
{"Phenylalanine", "F", "TTT, TTC"},
{"Methionine", "M", "ATG"},
{"Cysteine", "C", "TGT, TGC"},
{"Alanine", "A", "GCT, GCC, GCA, GCG"},
{"Proline", "P", "CCT, CCC, CCA,CCG "},
{"Threonine", "T", "ACT, ACC, ACA, ACG"},
{"Serine", "S", "TCT, TCC, TCA, TCG, AGT, AGC"},
{"Tyrosine", "Y", "TAT, TAC"},
{"Tryptophan", "W", "TGG"},
{"Glutamine", "Q", "CAA, CAG"},
{"Aspargine","N" "AAT, AAC"},
{"Histidine", "H", "CAT, CAC"},
{"Glutamic acid", "E", "GAA, GAG"},
{"Aspartic acid", "D", "GAT, GAC"},
{"Lysine", "K", "AAA, AAG"},
{"Arginine", "R", "CGT, CGC, CGA, CGG, AGA, AGG"},
{"Stop codons", "Stop", "AA, TAG, TGA"}
};
int count = 0;
printf("Enter the sequence: ");
gets(sequence);
char *input_string = sequence;
char *tmp_str = input_string;
int k;
char *pch;
while (*input_string != ' ')
{
char string_3l[4] = {' '};
strncpy(string_3l, input_string, 3);
printf("n-----------%s & %s----------", string_3l, tmp_str );
for(k=0;k<20;k++)
{
//printf("@REAL - %s", c1[0].dna_codon);
printf("@ %s", c1[k].dna_codon);
int x;
x = c1[k].dna_codon;
pch = strtok(x, ",");
while (pch != NULL)
{
printf("n%d : %s with %s", k, string_3l, pch);
count=strcmp(string_3l, pch);
if(count==0)
{
strcat(short_slc, c1[k].slc);
printf("n==>%s", short_slc);
}
pch = strtok (NULL, " ,.-");
}
}
input_string = input_string+3;
}
printf("nProtien sequence is : %sn", short_slc);
}
输入:
TAGTAG
输出:
如果您看到
printf("n-----------%s & %s----------", string_3l, tmp_str );
在两种迭代中,我们都发现结构中定义的值降低了。
我想知道为什么结构会减少它或我的错误?因为我被困在这里
所需的输出:
StopStop
strtok()
必须仅在字符串的重复副本上使用,因为它覆盖了"分界符"。在必要时使用' 0'生成令牌。
下面的代码将切碎字符串:
x = c1[k].dna_codon;
pch = strtok(x, ",");
例如:
String = "CTT, CTC, CTA, CTG, TTA, TTG"
第一次strtok()
调用A' 0'overwrites','
String = "CTT"