需要帮助调试段错误
我不知道我的代码有什么问题。它正在正确编译，但在输出时显示第二张图像。输出告诉我代码有什么问题，它必须提供第一张图所示的输出。
代码必须像这样运行：
但显示如下：
我不知道哪个块有什么问题。这是用于检查输入字符串是否将被接受的代码。
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
char *arr[1][3]= {{"aABb","c","d"}};
char input[15]= {'\0'};
char temp[15]= {'\0'};
char stack[15]= {'$','S'};
int ip=0;
int ct=0;
int top,i;
char x;
clrscr();
printf("\t\t\t Predictive parser\t\t\t\n");
printf("___________________________________________________\n\n");
printf("The grammar is :\n");
printf("\t\tS-->aABb\n");
printf("\t\tA-->c\n");
printf("\t\tB-->d\n\n");
printf("You have follow some rules\n");
printf("The string must end with $\n");
printf("Enter the String : ");
gets(input);
top=(strlen(stack))-1;
x=stack[top];
printf("_______________________________________________________\n");
printf("stack\t\t\t input\t\t\t production\n");
printf("________________________________________________________\n");
while(x!='\0')
{
if(x=='S'&&input[ct]=='a')
{
printf("\n");
for(i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t");
for(i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\tderivation using S-->aABb\n");
stack[top]='\0';
strcpy(temp,strrev(arr[0][0]));
strcat(stack,temp);
top=strlen(stack)-1;
}
else if(x=='A'&&input[ct]=='c')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
stack[top]='\0';
strcpy(temp,strrev(arr[0][1]));
strcat(stack,temp);
top=strlen(stack)-1;
printf("\t\t derivation usingA-->c\n");
}
else if(x=='B'&&input[ct]=='d')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
stack[top]='\0';
strcpy(temp,strrev(arr[0][2]));
strcat(stack,temp);
top=strlen(stack)-1;
printf("\t\t derivation using B-->d\n");
}
else if(x=='a'&&input[ct]=='a')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\t popping a from the stack");
input[ct]=' ';
ct++;
stack[top]='\0';
top--;
}
else if(x=='c'&&input[ct]=='c')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\t popping c from the stack");
input[ct]=' ';
ct++;
stack[top]='\0';
top--;
}
else if(x=='d'&&input[ct]=='d')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\t popping d from the stack");
input[ct]=' ';
ct++;
stack[top]='\0';
top--;
}
else if(x=='b'&&input[ct]=='b')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\t popping b from the stack");
input[ct]=' ';
ct++;
stack[top]='\0';
top--;
}
else if(x=='$'&&input[ct]=='$')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
stack[top]='\0';
printf("\t\tSuccessfull\n");
}
else
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
stack[top]='\0';
printf("\t\tUnsuccessfull\n");
getch();
exit(1);
}
x=stack[top];
}
printf("\n____________________ Exit program__________________");
getch();
}
1 个答案:
答案 0 :(得分：0)
怀疑是conio.h的原因，而其他过时的标头是这个。
#include<stdio.h>
#include <stdlib.h>
// #include<conio.h>
#include<string.h>
char* strrev(char *str)
{
char *p1;
int i, j;
//puts("Reversing");
//puts(str);
p1 = (char*)malloc(strlen(str) * sizeof(char));
for( i = strlen(str)-1, j =0; i >=0; i--, j++) {
p1[j] = str[i];
}
p1[j] = '\0';
puts(p1);
//puts("Done");
return p1;
}
void main()
{
char *arr[1][3]= {{"aABb","c","d"}};
char input[15]= {'\0'};
char temp[15]= {'\0'};
char stack[15]= {'$','S'};
int ip=0;
int ct=0;
int top,i;
char x;
printf("\t\t\t Predictive parser\t\t\t\n");
printf("___________________________________________________\n\n");
printf("The grammar is :\n");
printf("\t\tS-->aABb\n");
printf("\t\tA-->c\n");
printf("\t\tB-->d\n\n");
printf("You have follow some rules\n");
printf("The string must end with $\n");
printf("Enter the String : ");
gets(input);
top=(strlen(stack))-1;
x=stack[top];
printf("_______________________________________________________\n");
printf("stack\t\t\t input\t\t\t production\n");
printf("________________________________________________________\n");
while(x!='\0')
{
if(x=='S'&&input[ct]=='a')
{
printf("\n");
for(i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t");
for(i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\tderivation using S-->aABb\n");
stack[top]='\0';
strcpy(temp,strrev(arr[0][0]));
strcat(stack,temp);
top=strlen(stack)-1;
}
else if(x=='A'&&input[ct]=='c')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
stack[top]='\0';
strcpy(temp,strrev(arr[0][1]));
strcat(stack,temp);
top=strlen(stack)-1;
printf("\t\t derivation usingA-->c\n");
}
else if(x=='B'&&input[ct]=='d')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
stack[top]='\0';
strcpy(temp,strrev(arr[0][2]));
strcat(stack,temp);
top=strlen(stack)-1;
printf("\t\t derivation using B-->d\n");
}
else if(x=='a'&&input[ct]=='a')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\t popping a from the stack");
input[ct]=' ';
ct++;
stack[top]='\0';
top--;
}
else if(x=='c'&&input[ct]=='c')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\t popping c from the stack");
input[ct]=' ';
ct++;
stack[top]='\0';
top--;
}
else if(x=='d'&&input[ct]=='d')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\t popping d from the stack");
input[ct]=' ';
ct++;
stack[top]='\0';
top--;
}
else if(x=='b'&&input[ct]=='b')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
printf("\t\t popping b from the stack");
input[ct]=' ';
ct++;
stack[top]='\0';
top--;
}
else if(x=='$'&&input[ct]=='$')
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
stack[top]='\0';
printf("\t\tSuccessfull\n");
}
else
{
printf("\n");
for (i=0; i<=strlen(stack); i++)
printf("%c",stack[i]);
printf("\t\t\t ");
for (i=ip; i<6; i++)
printf("%c",input[i]);
stack[top]='\0';
printf("\t\tUnsuccessfull\n");
getchar();
exit(1);
}
x=stack[top];
}
printf("\n____________________ Exit program__________________");
getchar();
}
上面是您的代码，做了一些微小的修改
- 使用了自定义
strrev，这是一种非常幼稚的实现，具有不良的内存实践，但是可以正常工作。 - 摆脱
conio - 为
stdlib.h添加了malloc - 摆脱
getch的支持，getchar - 摆脱
clrscr()
没有人知道conio和旧的Borland标头中有什么错误。他们可能错误地重新定义了一些std函数。
当我用输入acdb$运行以上命令时，我得到了
Predictive parser
___________________________________________________
The grammar is :
S-->aABb
A-->c
B-->d
You have follow some rules
The string must end with $
Enter the String : acdb$
_______________________________________________________
stack input production
________________________________________________________
$S acdb$ derivation using S-->aABb
bBAa
$bBAa acdb$ popping a from the stack
$bBA cdb$c
derivation usingA-->c
$bBc cdb$ popping c from the stack
$bB db$d
derivation using B-->d
$bd db$ popping d from the stack
$b b$ popping b from the stack
$ $ Successfull
____________________ Exit program__________________
开始使用gcc / clang并使用gdb / lldb调试这些程序。
here是不错的指南。
如果使用这些选项进行编译，则可以看到您的程序充斥着不良的编程模式
gcc -Werror -Wall -Wextra -pedantic parser.c -o parser
parser.c: In function ‘strrev’:
parser.c:9:4: error: C++ style comments are not allowed in ISO C90 [-Werror]
parser.c:9:4: error: (this will be reported only once per input file) [-Werror]
parser.c:11:4: error: implicit declaration of function ‘malloc’ [-Werror=implicit-function-declaration]
parser.c:11:16: error: incompatible implicit declaration of built-in function ‘malloc’ [-Werror]
parser.c: At top level:
parser.c:21:6: error: return type of ‘main’ is not ‘int’ [-Werror=main]
parser.c: In function ‘main’:
parser.c:52:23: error: comparison between signed and unsigned integer expressions [-Werror=sign-compare]
parser.c:66:24: error: comparison between signed and unsigned integer expressions [-Werror=sign-compare]
parser.c:80:24: error: comparison between signed and unsigned integer expressions [-Werror=sign-compare]
parser.c:94:24: error: comparison between signed and unsigned integer expressions [-Werror=sign-compare]
parser.c:108:24: error: comparison between signed and unsigned integer expressions [-Werror=sign-compare]
parser.c:122:24: error: comparison between signed and unsigned integer expressions [-Werror=sign-compare]
parser.c:136:24: error: comparison between signed and unsigned integer expressions [-Werror=sign-compare]
parser.c:150:24: error: comparison between signed and unsigned integer expressions [-Werror=sign-compare]
parser.c:161:24: error: comparison between signed and unsigned integer expressions [-Werror=sign-compare]
parser.c:169:13: error: implicit declaration of function ‘exit’ [-Werror=implicit-function-declaration]
parser.c:169:13: error: incompatible implicit declaration of built-in function ‘exit’ [-Werror]
cc1: all warnings being treated as errors
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