malloc失败搞不定了 - 编程论坛

#2

不会游泳的虾2022-03-08 09:07

修改如下，供参考：

程序代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static int argnum;
static char** argstrs;
int main(int argc, char* argv[])
{
    argnum = argc;
    argstrs = (char**)malloc((argc - 1) * sizeof(char*));
    int k;
    for (k = 1; k < argc; k++) {
        int j = k - 1;
        char* str = argv[k];
        int slen = strlen(str);
        printf("%d: %s\t\t%d\n", k, str, slen);
        char* nstr = (char*)malloc(slen * sizeof(char));//传4个参数，最后一次此行会分配内存失败，不知道为什么
        //printf("%s\n", nstr);
        strcpy(nstr, str);
        argstrs[j] = (char*)malloc(slen * sizeof(char));//修改
        *(argstrs + j) = nstr;
        printf("%s\n", argstrs[j]);
    }
    return 0;
}

#3

rjsp2022-03-08 12:35

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

static int argnum;
static char** argstrs;

int main( int argc, char* argv[] )
{
    argnum = argc;
    argstrs = malloc( (argc-1)*sizeof(char*) );

    for( int k=1; k<argc; ++k )
    {
        char* str = argv[k];
        size_t slen = strlen(str);
        printf( "argv[%d]: %s\t\t,length=%zu\n", k, str, slen );

        char* nstr = malloc( (slen+1)*sizeof(char) );
        strcpy( nstr, str );
        printf( "%s\n", nstr );
        argstrs[k-1] = nstr;
    }
}

#4

udefine2022-03-08 14:54

确实是少分配了一字节内存，slen+1后就好了