Question

所以当我尝试调试我的程序时，我遇到了这些警告，我不知道如何处理。这是编译器给我的内容：

translate.c: In function 'main':

translate.c:65: warning: passing argument 2 of 'strncmp' makes pointer from integer without a cast 

/usr/include/string.h:146: note: expected 'const char *' but argument is of type 'char'

translate.c:67: warning: passing argument 1 of 'strcpy' makes pointer from integer without a cast

/usr/include/string.h:128: note: expected 'char * __restrict__' but argument is of type 'char'

translate.c:88: warning: passing argument 1 of 'strcpy' from incompatible pointer type

/usr/include/string.h:128: note: expected 'char * __restrict__' but argument is of type 'struct node *'

translate.c:100: warning: format '%s' expects type 'char *', but argument 3 has type 'struct node *'

这是我的代码，我会评论它所指的区域：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

typedef struct node{
    char seq[300];
    struct node* next;
    } NODE;


int
main(int argc, char* argv[]){

    int i, j=0;
    FILE *fin, *fout, *fop;
    char code1[100], code2[100], prot1[2], prot2[4];
    NODE *current, *first, *prev;

    fin = fopen( argv[1], "r");
    fout = fopen( argv[2], "w");
    fop = fopen("codeoflife.txt", "r");

    code2[0] = '\0';
    current = first = malloc (sizeof (NODE));

    while( fscanf( fin, "%s", current -> seq) != EOF) {

            for (i = 0; i < 300; i++){
                    if (current->seq[i] == 'a')
                            current->seq[i] = 'A';
                    else if (current->seq[i] == 't')
                            current->seq[i] = 'T';
                    else if(current->seq[i] == 'g')
                            current->seq[i] = 'G';
                    else if(current->seq[i] == 'c')
                            current->seq[i] = 'C';
            }

            if ( (current -> next = malloc ( sizeof(NODE) ) ) == NULL){
                    fprintf(fout, "Out of memory\nCan't add more DNA sequences\n");
                    return EXIT_FAILURE;
            }
            prev = current;
            current = current -> next;

    }

    fclose(fin);
    current = first;


    while(current->next != NULL){
            for( i = 0; i < 300; i++){
                    if( current->seq[i] == 'A')
                            if( current->seq[i+1] == 'G')
                                    if( current->seq[i+2] =='T'){
                                            i = i+3;
                                            j = 0;
                                            code1[j] = 'M';
                                            while(fscanf(fop, "%s%s", prot1, prot2) != EOF){
                                                    if(strcmp(prot2, "TAA") == 0 || strcmp(prot2, "TAG") == 0 || strcmp(prot2, "TGA") == 0){
                                                            fclose(fop);
                                                            break;
                                                    }
                        /* line 65 */                           if(strncmp(prot2, current->seq[i], 3) == 0){
                                                            j++;
                        /* line 67*/                                    strcpy(code1[j], prot1);
                                                            fclose(fop);
                                                            i = i + 3;
                                                    }
                                            }

                                            if(strcmp(code1, code2) > 0)
                                                    strcpy(code2, code1);
                                            //else if (strcmp(code1, code2) < 0)

                                    }


            }

            if(strcmp(prot2, "TAA") != 0 && strcmp(prot2, "TAG") != 0 && strcmp(prot2, "TGA") != 0){
                    strcpy ( current->seq, "None");
                    current = current->next;
                    continue;
            }

   /* line 88 */         strcpy(current->next, code2);

            current = current->next;

            code2[0] = '\0';
    }

    free(current);
    prev->next = NULL;
    current = first;

    while(current->next != NULL){
 /* line 100 */           fprintf( fout, "\n%s\n", current->next);
            current = current->next;
    }

    return 0;
   }

请帮助我理解这一点。

Answer 1

current->seq为char[]（即const char *），因此current->seq[i]为char。

如果您打算使用strncmp来比较以current->seq字符开头的i内容，请将其调用为strncmp(xxx, current->seq+i, N)（例如strncmp(prot2, current->seq+i, 3)

Answer 2

或许，您需要获取字符串字符的地址，而不是字符，并进行某种类型的转换：

65: strncmp(prot2, &current->seq[i], 3)
67: strcpy(&code1[j], prot1)
88: strcpy((char *)current->next, code2)
100: fprintf( fout, "\n%s\n", (char *)current->next);

调试 - 字符串问题

2 个答案: